MCQ 11 Mark
Value of $\sum^{\infty}_{\text{k}=1}\sum^{\text{k}}_{\text{r}=0}\frac{1}{3^{\text{k}}}\big({^\text{k}}\text{C}_{\text{r}}\big)$ is:
- ✓
$2$
- B
$\frac{2}{3}$
- C
$\frac{1}{3}$
- D
$\text{None of these}$
Answer$\sum\frac{1}{3^{\text{k}}}{^\text{k}}\text{C}_{\text{r}}$
$=\frac{1}{3^{\text{k}}}\sum{^\text{k}}\text{C}_{\text{r}}$
$=\frac{2^{\text{k}}}{3^{\text{k}}}$
This is a G.P
Therefore, the sum of the series will be
$\text{S}=\frac{\frac{2}{3}}{1-\frac{20}{3}}=2$
View full question & answer→MCQ 21 Mark
If the rth term in the expansion of $\Big(\frac{\text{x}}{3}-\frac{2}{\text{x}^{2}}\Big)^{10}$ contains $x^4$, then r is equal to:
View full question & answer→MCQ 31 Mark
Sum of the coefficients of $(1-x)^{25}$ is:
Answer
- 0
Solution:
$(1-x)^{25}=1-{ }^{25} C_{1 x}+{ }^{25} C_2 x^2-{ }^{25} C_3 x^3+{ }^{25} C_4 x^4-{ }^{25} C_5 x^5 \ldots-{ }^{25} C_{25} x^{25}$
Putting $x=1$, we get
$0=1-{ }^{25} C_1+{ }^{25} C_2-{ }^{25} C_3+{ }^{25} C_4-{ }^{25} C_5 \ldots-{ }^{25} C_{25}$
Hence, sum of coefficients is 0 . View full question & answer→MCQ 41 Mark
The number of integral terms in the expansion of $\Big(3^{\frac{1}{8}}+5^{\frac{1}{4}}\Big)^{1024}$ is:
AnswerTotal number of integral term will be
$\frac{1024}{\text{L}.\text{C}.\text{M}(4,8)}+1$
$=\frac{1024}{8}+1$
$=128+1$
$=129.$
Hence there are total 129 integral terms.
View full question & answer→MCQ 51 Mark
The $4^{th}$ term in the expansion of $\Big(\sqrt{\text{x}}+\frac{1}{\text{x}}\Big)^{12}$ is:
AnswerCorrect option: B. $220\text{x}^{\frac{3}{2}}$
- $220\text{x}^{\frac{3}{2}}$
Solution:
Expansion is $\Big(\sqrt{\text{x}}+\frac{1}{\text{x}}\Big)^{12}$
$\text{T}_{\text{r}+1}=12_{\text{C}\text{r}}\big(\frac{1}{\text{x}}\big)^{\text{r}}\cdot(\sqrt{\text{x}})^{12-\text{r}}=12_{\text{C}\text{r}}\cdot\text{x}^{6-1.5\text{r}}$
$4^{th}$ term is $\text{T}_4=12_{\text{C}3}\cdot\text{x}^{6-1.5\times3}=220.\text{x}^{\frac{3}{2}}$ View full question & answer→MCQ 61 Mark
The expansion $\Big(\text{x}-\frac{\text{x}^{2}}{2}\Big)^{40}$ is a polynomial of $n^{th}$ degree in x, then n =
Answer
- 80
Solution:
$T_{r+1}={ }^{40} C_r x^{40-r} x^{2 r} 2^{-r}$
The power of $x=40+r$
Highest power of $x$ occurs when $r=40$ (last term)
Hence, highest power of $x$ is 80 .
Hence, the polynomial is of degree 80 . View full question & answer→MCQ 71 Mark
The approximate value of $(7.995)^{\frac{1}{3}}$ correct to 4 decimal places is:
View full question & answer→MCQ 81 Mark
Sum of the coefficients of $(1+x)^n$ is always a:
Answer
- An integer
Solution:
To determine the sum of coefficients, we substitute $\mathrm{x}=1$ in the above expression.
Thus sum of coefficients $=(1+1)^n=2^n$
Hence, its a positive integer. View full question & answer→MCQ 91 Mark
The number of terms with integral coefficients in the expansion of $\Big(7^{\frac{1}{3}}+5^{\frac{1}{2}}\cdot\text{x}\Big)^{600}$ is:
AnswerThe number of integral terms will be
$1+\frac{100}{\text{L}.\text{C}.\text{M}(2,3)}$
$=1+\frac{600}{6}$
$=1+100$
$=101$
View full question & answer→MCQ 101 Mark
The total number of terms in the expansion of $(x+a)^{100}+(x-a)^{100}$ after simplification is:
Answer
- 51
Solution:
In the above binomial expansion, the terms at the even places will get eliminated, and we would be left with twice the sum of the terms at odd places.
Hence there will be
$\frac{\text{n}}{2}+1$
$=\frac{100}{2}+1$
$=51\text{terms}$ View full question & answer→MCQ 111 Mark
In the expansion of $\Big(\frac{3\sqrt{\text{x}}}{3}-\frac{\sqrt{3}}{\text{x}}\Big)10,$ x > 0, the constant term is:
View full question & answer→MCQ 121 Mark
If in the expansion of $\Big(\text{x}-\frac{1}{3\text{x}^{3}}\Big)^{9},$ the term independent of $x$ is:
- A
$\text{T}_{3}$
- ✓
$\text{T}_{4}$
- C
$\text{T}_{5}$
- D
AnswerCorrect option: B. $\text{T}_{4}$
- $\text{T}_{4}$
Solution:
Suppose $T_{r+1}$ is the term in the given expression that is independent of x.
Thus, we have
$\text{T}_{\text{r}+1}={^\text{9}}\text{C}_{\text{r}}\ \text{x}^{9-\text{r}}\Big(\frac{-1}{3\text{x}^{2}}\Big)^{\text{r}}$
$=(-1)^{\text{r}}\ {^\text{9}}\text{C}_{\text{r}}\frac{1}{3^{\text{r}}}\ \text{x}^{9-\text{r}-2\text{r}}$
For this term to be independent of x, we must have
$9-3\text{r}=0$
$\Rightarrow\text{r}=3$
Hence, the required term is the $4^{th}$ term i.e. $\text{T}_{4}$ View full question & answer→MCQ 131 Mark
The number of terms in the expansion of $\left(a_1+b_1\right)\left(a_2+b_2\right) \ldots . . .\left(a_n+b_n\right)$.
- ✓
$2^n$
- B
$3^n$
- C
$3^{2 n}$
- D
$2^{2 n}$
Answer
- $2^n$
Solution:
Each bracket in the above expansion contains 2 elements
Therefore
2 brackets will have $2^2=4$ elements
3 brackets will have $2^3=8$ elements
4 brackets will have $2^4=16$ elements
:
:
$n$ brackets will have $2^n$ elements
Hence, there will be $2^n$ elements View full question & answer→MCQ 141 Mark
The number of terms with integral coefficients in the expansion of $\Big(17^{\frac{1}{3}}+35^{\frac{1}{2}}\text{x}\Big)^{600}$ is:
Answer
- 101
Solution:
The general term $T_{r+1}$ in the given expansion is given by ${^\text{600}}\text{C}_{\text{r}}(17^{\frac{1}{3}})^{600-\text{r}}\Big(35^{\frac{1}{2}}\text{x}\Big)^{\text{r}}$
$={^\text{600}}\text{C}_{\text{r}}(17^{\frac{1}{3}})^{200-\frac{\text{r}}{3}}\times\Big(35^{\frac{\text{r}}{2}}\text{x}\Big)^{\text{r}}$
Now, $T_{r+1}$ is an integer if $\frac{\text{r}}{2}$ and $\frac{\text{r}}{2}$ are integers for all $0\leq\text{r}\leq600$
Thus, we have
r = 0, 6, 12, ....600
Since, It is an A.P
So, $600 = 0 + (\text{n} - 1)6$
$\Rightarrow \text{n}=101$
Hence, there are 101 terms with integral coefficients. View full question & answer→MCQ 151 Mark
If the $(n+1)$ numbers $a, b, c, d, \ldots a, b, c, d, \ldots$ be all different and each of them a prime number, then the number of different factors (other than 1 ) of $a^m$. b.c.d.... is:
- A
$m-2^n$
- B
$(m+1) 2^n$
- ✓
$(m+1) 2^n-1$
- D
AnswerCorrect option: C. $(m+1) 2^n-1$
- $(m+1) 2^n-1$
Solution:
No. of elements in b.c.d... $=\mathrm{n}$
Choose $a^k$, where $k \in 0,0,1,2, \ldots m$ at a time $=(m+1)$ for every $k$, no. possible factors from $n$ elements
$={ }^n C_0+{ }^n C_1+{ }^n C_2+\ldots . .+{ }^n C_n$
$=2^n$
Total factors $=$ No. of possible powers of a $\times$ every $k$, no. possible factors from $n$ elements
$=(m+1) 2^n$
If factor 1 is excluded then
$(m+1) 2 n-1$ View full question & answer→MCQ 161 Mark
$\sum\limits^{16}_{\text{r}+2}{^{16}}\text{C}_\text{r}=$
- A
$2^{15}-15$
- B
$2^{16}-16$
- ✓
$2^{16}-17$
- D
$2^{17}-17$
AnswerCorrect option: C. $2^{16}-17$
- $2^{16}-17$
Solution:
Consider given the binomial expression,
$\sum\limits^{16}_{\text{r}+2}{^{16}}\text{C}_\text{r}={^{16}}\text{C}_{2}+{^{16}}\text{C}_{3}+{^{16}}\text{C}_{3}+\ .....\ {^{16}}\text{C}_{16}$
$=2^{16}-17$
Hence, this is the answer. View full question & answer→MCQ 171 Mark
If in the expansion of $(1+\text{x})^{20},$ the coefficients of rth and (r + 4) terms are equal, then r is equal to:
AnswerCoefficients of the rth and (r + 4)th terms in the given expansion are ${^\text{20}}\text{C}_{\text{r}-1}$ and ${^\text{20}}\text{C}_{\text{r}}.$
Here,
${^\text{20}}\text{C}_{\text{r}-1}={^\text{20}}\text{C}_{\text{r}+3}$
$\Rightarrow \text{r}-1+\text{r}+3=20$
$\Rightarrow \text{r}=2$ or $2\text{r}=18$
$\Rightarrow \text{r}=9$
View full question & answer→MCQ 181 Mark
If the number of terms in $\Big(\text{x}+1+\frac{1}{\text{x}}\Big)^\text{n}(\text{n}\in\text{I}^{+})$ is 401, then n is greater than.
AnswerIf we substitute 2 in place of 1 in the above expression, we get
$\Big(\text{x}+2+\frac{1}{\text{x}}\Big)^\text{n}$
$\Big(\Big(\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big)\Big)^{2\text{n}}$
Since the total number of terms is 401
2n + 1 = 401
2n = 400
n = 200
However the actual question is
$\Big(\text{x}+1+\frac{1}{\text{x}}\Big)^\text{n}$
Hence n is greater than 199.
View full question & answer→MCQ 191 Mark
If $(1-\text{x}^{2})^{\text{n}}=\sum_{\text{r}=0}^{\text{n}}\text{a}_{\text{r}}\text{x}^{\text{r}}(1-\text{x})^{2\text{n}-\text{r}},$ then $a_r$ is equal to:
- A
${ }^n C_r$
- ✓
${ }^n \mathrm{C}_{\mathrm{r}} 3^{\mathrm{r}}$
- C
${ }^{2 n} \mathrm{C}_{\mathrm{r}}$
- D
${ }^n \mathrm{C}_{\mathrm{r}} 2^{\mathrm{r}}$
AnswerCorrect option: B. ${ }^n \mathrm{C}_{\mathrm{r}} 3^{\mathrm{r}}$
- ${ }^n \mathrm{C}_{\mathrm{r}} 3^{\mathrm{r}}$
Solution:
$(1-\text{x})^{\text{n}}(1+\text{x})^{\text{n}}=\sum_{\text{r}=0}^{\text{n}}\text{a}_{\text{r}}\text{x}^{\text{r}}(1-\text{x})^{\text{n}}(1-\text{x})^{\text{n}-\text{r}}$
$\Rightarrow(1-\text{x}+2\text{x})^{\text{n}}=\sum_{\text{r}=0}^{\text{n}}\text{a}_{\text{r}}\text{x}^{\text{r}}(1-\text{x})^{\text{n}-\text{r}}$
$\Rightarrow\sum_{\text{r}=0}^{\text{n}}{^\text{n}}\text{C}_{\text{r}}(1-\text{x})^{\text{n}-\text{r}}(2\text{x})^{\text{r}}$
$=\sum_{\text{r}=0}^{\text{n}}\text{a}_{\text{r}}\text{x}^{\text{r}}(1-\text{x})^{\text{n}-\text{r}}$
Comparing general term, we get $\text{a}^{\text{r}}={^\text{n}}\text{C}_{\text{r}}2^{\text{r}}$ View full question & answer→MCQ 201 Mark
Number of irrational terms in the binomial expansion of $\Big(3^{\frac{1}{5}}+7^{\frac{1}{3}}\Big)^{100}$ is:
AnswerAs n = 100 hence there are 101 terms
The genral term for expansion is given as $\text{T}_{\text{r}+1}={^{100}}\text{C}_{\text{r}}\cdot3\frac{100-\text{r}}{5}\cdot7\frac{\text{r}}{3}$
So, we need to find such values of r for which $\frac{100-\text{r}}{5}$ and $\frac{\text{r}}{3}$ is a natural number.
For such value we find value of r i.e r = 0, 15, 30, 45, 60, 75, 90
hence there are 7 values for which it is rational.
So there are 101 - 7 = 94 irrational terms.
View full question & answer→MCQ 211 Mark
If the sum of the binomial coefficients of the expansion $\Big(2\text{x}+\frac{1}{\text{x}}\Big)^{\text{n}}$ is equal to 256, then the term independent of x is:
Answer
- 1120
Solution:
Suppose $(r + 1)^{th}$ term in the given expansion is independent of x.
Then, we have
$\text{T}_{\text{r}+1}={^\text{n}}\text{C}_{\text{r}}(2\text{x})^{\text{n}-\text{r}}\Big(\frac{1}{\text{x}}\Big)^{\text{r}}$
$={^\text{n}}\text{C}_{\text{r}}(2)^{\text{n}-\text{r}}\text{x}^{\text{n}-2\text{r}}$
For this term to be independent of x, we must have
$\text{n}-2\text{r}=0$
$\Rightarrow \text{r}=\frac{\text{n}}{2}$
$\therefore$ Required term $={^\text{n}}\text{C}_{\frac{\text{n}}{2}}\ 2^{\text{n}-\frac{\text{n}}{2}}=\frac{\text{n!}}{\big[(\frac{\text{n}}{2})\big]}\ 2^{\frac{\text{n}}{2}}$
We know,
Sum of the given expansion = 256
Thus, we have
$2^{\text{n}}.1^{\text{n}}=256$
$\Rightarrow \text{n}=8$
$\therefore$ Required term $=\frac{8!}{(4)!(4)!}2^{4}=1120$ View full question & answer→MCQ 221 Mark
Which of the following is the highest value?
- A
$12^9$
- B
$10^{11}$
- C
$11^{10}$
- ✓
Answer
- All option are right
Solution:
We know a prime x > 6 can be always written as 6k + 1 or 6k - 1, when k is an integer.
Option A ⟶ 889 = 6 × 148 + 1
Option B ⟶ 997 = 6 × 166 + 1
Option C ⟶ 899 = 6 × 133 + 1
Option D ⟶ 1147 = 6 × 191 + 1
$\therefore$ All the options are correct. View full question & answer→MCQ 231 Mark
[AS 1] If $\text{A}=\frac{1}{3}\text{B}$ and $\text{B}=\frac{1}{2}\text{C},$ then A : B : C = ..
Answer$\text{A}=\frac{\text{B}}{3}....(1)$
$\text{B}=\frac{\text{C}}{2}$
$\Rightarrow\text{C}=2\text{B}.....(2)$
From (1) and (2),
$\text{A}:\text{B}:\text{C}=\frac{\text{B}}{3}:\text{B}:2\text{B}$
$=\frac{1}{3}:1:2$
$=1:3:6$
View full question & answer→MCQ 241 Mark
If the co - efficient of x in $\Big(\text{x}^2+\frac{\lambda}{\text{x}}\Big)^5$is 270, then $ \lambda=$
AnswerNow, the co - efficient of x in $\Big(\text{x}^2+\frac{\lambda}{\text{x}}\Big)^5$is ${^6}\text{C}_{3}\cdot(\lambda)^3$or $10(\lambda)^3.$
According to the problem $10(\lambda)^{3}=270,$ or $(\lambda)=3.$
View full question & answer→MCQ 251 Mark
If the coefficients of $2^{\text {nd}}, 3^{\text {rd}}$ and $4^{\text {th}}$ terms in the expansion of $(1+\text{x})^{\text{n}}, \text{n}\in\text{N}$ are in A.P. then n =
Answer
- 7
Solution:
Coefficients of $2^{\text {nd }}, 3^{\text {rd }}$ and $4^{\text {th }}$ terms in the expansion of are ${^\text{n}}\text{C}_{\text{1}},{^\text{n}}\text{C}_{\text{2}}, {^\text{n}}\text{C}_{\text{3}}.$
we have,
$2\times{^\text{n}}\text{C}_{\text{2}}={^\text{n}}\text{C}_{\text{1}}+{^\text{n}}\text{C}_{\text{3}}$
Dividing both sides by ${^\text{n}}\text{C}_{\text{2}}$, we get
$2=\frac{{^\text{n}}\text{C}_{\text{1}}}{{^\text{n}}\text{C}_{\text{2}}}+\frac{{^\text{n}}\text{C}_{\text{3}}}{{^\text{n}}\text{C}_{\text{2}}}$
$\Rightarrow 2=\frac{2}{\text{n}-1}+\frac{\text{n}-2}{3}$
$\Rightarrow 6\text{n}-6=6+\text{n}^{2}+2-3\text{n}$
$\Rightarrow \text{n}^{2}-9\text{n}+14=0$
$\Rightarrow \text{n}=7$ View full question & answer→MCQ 261 Mark
If the sum of odd numbered terms and the sum of even numbered terms in the expansion of $(\text{x}+\text{a})^{\text{n}}$ are A and B respectively, then the value of $(\text{x}^{2}-\text{a}^{2})^{\text{n}}$ is:
- ✓
$A ^2- B ^2$
- B
$A^2+B^2$
- C
- D
AnswerCorrect option: A. $A ^2- B ^2$
- $A ^2- B ^2$
Solution:
If A and B denote respectively the sums of odd terms and even terms in the expansion $(\text{x}+\text{a})^{\text{n}}.$
Then,
$(\text{x}+\text{a})^{\text{n}}=\text{A}+\text{B}\ ...(\text{i})$
$(\text{x}-\text{a})^{\text{n}}=\text{A}-\text{B}\ ...(\text{ii})$
Multplying both the equations we get,
$(\text{x}+\text{a})^{\text{n}}(\text{x}-\text{a})^{\text{n}}=\text{A}^{2}-\text{B}^{2}$
$\Rightarrow (\text{x}-\text{a})^{\text{n}}=\text{A}^{2}-\text{B}^{2}$ View full question & answer→MCQ 271 Mark
If n is the positive integer, then $2^{3n} - 7n - 1$ is divisible by.
Answer
- 49
Solution:
Given: $2^{3 n}-7 n-1$. It can also be written as $8^n-7 n-1$
Let $8^n-7 n-1=0$
So, $8^n=7 n+1$
$8^n=(1+7)^n$
By applying binomial theorem, we get
$8 n-1-7 n=49$ (or) $2^{3 n}-7 n-1=49$
Hence, $2^{3 n}-7 n-1$ is divisible by 49 . View full question & answer→MCQ 281 Mark
If $(1+a x)^n=1+8 x+24 x^2+\ldots .$. then $a \times n$ is:
Answer
- 8
Solution:
$(1+a x)^n=1+8 x+24 x^2+.......$
$\Rightarrow{ }^n C_0+{ }^n C_1(a x)+{ }^n C_2 \cdot(a x)^2+\ldots \ldots=1+8 x+24 a x^2 \ldots \ldots \ldots . .$
$\Rightarrow 1+(na) x+{ }^n C_2 \cdot(ax)^2+\ldots \ldots=1+8 x+24 a x^2 \ldots \ldots \ldots . .$
Comparing coefficient of $x$ in R.H.S to that in L.H.S.Thus $n \times a =8$ View full question & answer→MCQ 291 Mark
If in the expansion of $(\text{a}+\text{b})^{\text{n}}$ and $(\text{a}+\text{b})^{\text{n}}+3,$ the ratio of the coefficients of coefficients of second and third terms, and third and fourth terms respectively are equal, then n is:
AnswerCoefficients of the 2nd and 3rd terms in $\Big(\text{a}+\text{b}\Big)^{\text{n}}$ are ${^\text{n}}\text{C}_{\text{}1}$ and ${^\text{n}}\text{C}_{\text{}2}$
Coefficients of the 2nd and 3rd terms in $\Big(\text{a}+\text{b}\Big)^{\text{n}+3}$ are ${^\text{n+3}}\text{C}_{\text{}2}$ and ${^\text{n+3}}\text{C}_{\text{}3}$
Thus, we have
$\frac{{^\text{n}}\text{C}_{\text{}1}}{{^\text{n}}\text{C}_{\text{}2}}=\frac{{^\text{n+3}}\text{C}_{\text{}1}}{{^\text{n+3}}\text{C}_{\text{}3}}$
$\Rightarrow \frac{2}{\text{n}-1}=\frac{3}{\text{n}+1}$
$\Rightarrow 2\text{n}+2=3\text{n}-3$
$\Rightarrow \text{n}=5$
View full question & answer→MCQ 301 Mark
If in the binomial expansion of (1 + x)n where n is a natural number, the coefficients of the 5th, 6th and 7th terms are in A.P., then n is equal to:
MCQ 311 Mark
The coefficient of $x^5$ in the expansion of $(1+\text{x})^{21}+(1+\text{x})^{22}+...+(1+\text{x})^{30}$
- A
${^\text{51}}\text{C}_{\text{5}}$
- B
${^\text{9}}\text{C}_{\text{5}}$
- ✓
${^\text{31}}\text{C}_{\text{6}}-{^\text{21}}\text{C}_{\text{6}}$
- D
${^\text{30}}\text{C}_{\text{5}}+{^\text{20}}\text{C}_{\text{5}}$
AnswerCorrect option: C. ${^\text{31}}\text{C}_{\text{6}}-{^\text{21}}\text{C}_{\text{6}}$
- ${^\text{31}}\text{C}_{\text{6}}-{^\text{21}}\text{C}_{\text{6}}$
Solution:
we have,
$(1+\text{x})^{21}+(1+\text{x})^{22}+...+(1+\text{x})^{30}$
$=(1+\text{x})^{21}\Big[\frac{(1+\text{x})^{10}-1}{(1+\text{x})+1}\Big]$
$=\frac{1}{\text{x}}\Big[(1+\text{x})^{31}-(1+\text{x})^{21}\Big]$
Coefficient of $x^5$ in the given expansion = Coefficient of $x^5$ in $=\frac{1}{\text{x}}\Big[(1+\text{x})^{31}-(1+\text{x})^{21}\Big]$
= Coefficient of $x^6$ in $\Big[(1+\text{x})^{31}-(1+\text{x})^{21}\Big]$
$={^\text{31}}\text{C}_{\text{6}}-{^\text{21}}\text{C}_{\text{6}}$ View full question & answer→MCQ 321 Mark
The term without x in the expansion of $(2\text{x}-\frac{1}{2\text{x}^{2}}\Big)^{12}$ is:
Answer
- 7920
Solution:
Suppose the $(r + 1)^{th}$ term in the given expansion is independent of x.
Then, we have,
$\text{T}_{\text{r}+1}={^\text{12}}\text{C}_{\text{r}}(2\text{x})^{12-\text{r}}\Big(\frac{-1}{2\text{x}^{2}}\Big)^{\text{r}}$
$=(-1)^{\text{r}}\ {^\text{12}}\text{C}_{\text{r}}\ 2^{12-2\text{r}}\ \text{x}^{12-\text{r}-2\text{r}}$
For this term to be independent of x, we must have:
$=12-3\text{r}=0$
$\Rightarrow\text{r}=4 $
$\therefore$ Required term,
$(-1)^{4}\ {^\text{12}}\text{C}_{\text{4}}\ 2^{12-8}$
$=\frac{12\times11\times10\times9}{4\times3\times2}\times16$
$=7920$ View full question & answer→MCQ 331 Mark
The coefficient of $x^{-3}$ in the expansion of $\Big(\text{x}-\frac{\text{m}}{\text{x}}\Big)^{11}$ is:
- A
$-924 m^7$
- B
$-792 m^5$
- C
$-792 m^6$
- ✓
$-330 m^7$
AnswerCorrect option: D. $-330 m^7$
- $-330 m^7$
Solution:
Let $x^{-3}$ occur at $(r + 1)^{th}$ term in the given expansion.
Then, we have
$\text{T}_{r+1}={^\text{11}}\text{C}_{\text{r}}\ \text{x}^{11-\text{r}}\ \Big(\frac{-\text{m}}{\text{x}}\Big)^{\text{r}}$
$=(-1)^{\text{r}}\times{^\text{11}}\text{C}_{\text{r}}\ \text{m}^{\text{r}}\ \text{x}^{11-\text{r}-\text{r}}$
For this term to contain $x^{-3}$, we must have
$=11-2\text{r}=-3$
$\Rightarrow \text{r}=7$
Required coefficient $=(-1)^{7}\ {^\text{11}}\text{C}_{\text{7}}\ \text{m}^{7}$
$=-\frac{11\times10\times9\times8}{4\times3\times2}\ \text{m}^{7}$
$=-330\text{m}^{7}$ View full question & answer→MCQ 341 Mark
How many terms are there in the expansion of $\left(1+2 x+x^2\right)^{10}$?
Answer
- 21
Solution:
Now, $\left(1+2 x+x^2\right)^{10}=\left((1+x)^2\right)^{10}=(1+x)^{20}$
Now, the number of terms in the expansion of $(1+x)^n$ are $n+1$.
Thus, the number of terms in the expansion of $(1+x)^{20}$ will be $20+1=21$. View full question & answer→MCQ 351 Mark
The sum of the coefficients of all the even powers of $x$ in the expansion of $\left(2 x^2-3 x+1\right)^{11}$ is:
- A
$2.6^{10}$
- ✓
$3.6^{10}$
- C
$6^{11}$
- D
AnswerCorrect option: B. $3.6^{10}$
- $3.6^{10}$
Solution:
Given equation is $\left(2 x^2-3 x+1\right)^{11}$
$=(2 x-1)^{11}(x-1)^{11}$
$=(3)^{11} \cdot(2)^{11-1}$
$=3^{11} \cdot 2^{10}$
$=6^{10} \cdot 3$ View full question & answer→MCQ 361 Mark
The 4th term from the end in the expansion of $\Big(\frac{\text{x}^3}{2}-\frac{2}{\text{x}^{2}}\Big)^7$ is:
- A
$35 x$
- B
$70 x^2$
- C
$35 x^2$
- ✓
$70 x$
AnswerCorrect option: D. $70 x$
- $70 x$
Solution
For the above question
$T_{r+1}={ }^7 C_r x^{21-5r} 2^{2 r-7}$
For the fourth term, from the end $r =4$
$T_{5+1}={ }^7 C_4 x^1 2$
$=(35)(2) x$
$=70 x$ View full question & answer→MCQ 371 Mark
The coefficient of $x^4$ in the expansion of $(1 - 2x)^5$ is equal to:
Answer
- 80
Solution:
General term of $(1-2 x)^5$ is given by
$T_{r+1}={ }^5 C_r(-2 x)^r$
$={ }^5 C_r(-2) x^r$
For coefficient of $x^4$, power of $x=4$
$\therefore \mathrm{r}=4$
$\therefore \text { Coefficient pf } \mathrm{x}^4={ }^5 \mathrm{C}_4(-2)^4$
$=5 \times 16=80$ View full question & answer→MCQ 381 Mark
The number of non - zero terms in the expansion of $\big(1+3\sqrt{2}\text{x}\big)^{9}+\big(1-3\sqrt{2}\text{x}\big)^{9}$ is:
AnswerIn the expansion of $\big(1+3\sqrt{2}\text{x}\big)^{9}+\big(1-3\sqrt{2}\text{x}\big)^{9}$ 2nd, 4th, 6th, 8th and 10th terms get cancelled.
$\therefore$ Number of non - zero terms in $2\Big[{^9}\text{C}_{0}+{^9}\text{C}_{2}(3\sqrt{2}\text{x})^{2}+\ ...\ +{^9}\text{C}_{8}(3\sqrt{2}\text{x})^{8}\Big]$ is 5.
View full question & answer→MCQ 391 Mark
If n is even in the expansion of (a + b)n, the middle term is:
- A
$\text{n}^\text{th }\text{term}$
- B
$\big(\frac{\text{n}}{2}\big)^\text{th }\text{term}$
- C
$\big[\big(\frac{\text{n}}{2})-1\big]^\text{th }\text{term}$
- ✓
$\big[\big(\frac{\text{n}}{2})+1\big]^\text{th }\text{term}$
AnswerCorrect option: D. $\big[\big(\frac{\text{n}}{2})+1\big]^\text{th }\text{term}$
- $\big[\big(\frac{\text{n}}{2})+1\big]^\text{th }\text{term}$
Solution:
In general, if " $n$ " is the even in the expansion of $(a+b)^n$, then the number of terms will be odd. (i.e) $n+1$. Hence, the middle term of the expansion $(a+b)^n$ is $\left[\left(\frac{n}{2}\right)+1\right]^{\text {th }}$ term. View full question & answer→MCQ 401 Mark
The sum of the co - efficients of all odd degree terms in the expansion of$(\text{x}+\sqrt{\text{x}^3-1})^5+(\text{x}+\sqrt{\text{x}^3-1})^5(\text{x}>1)$ is:
Answer
- 2
Solution:
Sum of the coefficient of odd term is given by
$=2\left[{ }^5 \mathrm{C}_0+{ }^5 \mathrm{C}_2+{ }^5 \mathrm{C}_4\right]$
$=2[1+10-10+5-10+5]$
$=2(1+5+5-10)=2 .$ View full question & answer→MCQ 411 Mark
Constant term in the expansion of $\Big(\text{x}-\frac{1}{\text{x}}\Big)^{10}$ is:
Answer
- -252
Solution:
Suppose $(r + 1)^{th}$ term is the constant term in the given expansion.
Then, we have
$\text{T}_{\text{r}+1}={^\text{10}}\text{C}_{\text{r}}\ \text{x}^{10-\text{r}}\ \Big(\frac{\text{1}}{\text{x}}\Big)^{\text{r}}$
$={^\text{10}}\text{C}_{\text{r}} (-1)^{\text{r}}\ \text{x}^{10-\text{r}-\text{r}}$
For this term to be constant, we must have
$10-2\text{r}=0$
$\Rightarrow \text{r}=5$
Required term $={^\text{10}}\text{C}_{\text{5}}=-252$ View full question & answer→MCQ 421 Mark
${ }^{(2 n+1)} C_0-{ }^{(2 n+1)} C_1+{ }^{(2 n+1)} C_2-\ldots . . .^{2 n+1} C_{2 n}=$
Answer
- 1
Solution:
In some questions, substituting n = a positive number in both the question and the answer is the fastest way to achieve the correct option.
Although there is always alternate options like writing the general term and splitting it to a format that can be solved but that takes long in a limited time paper.
Try to put n = 1. In the end only option A remains.
View full question & answer→MCQ 431 Mark
If $\frac{\text{T}_{2}}{\text{T}_{3}}$ in the expansion of $(\text{a}+\text{b})^{\text{n}}$ and $\frac{\text{T}_{3}}{\text{T}_{4}}$ in the expansion of $(\text{a}+\text{b})^{\text{n}+3}$ are equal, then n =
AnswerIn the expansion $(\text{a}+\text{b})^{\text{n}},$ we have
$\frac{\text{T}_{2}}{\text{T}_{3}}=\frac{{^\text{n}}\text{C}_{\text{1}}\text{a}^{\text{n}-1}\times\text{b}^{1}}{{^\text{n}}\text{C}_{\text{2}}\text{a}^{\text{n}-2}\times\text{b}^{2}}$
In the expansion $(\text{a}+\text{b})^{\text{n}+3},$ we have
$\frac{\text{T}_{3}}{\text{T}_{4}}=\frac{{^\text{n+3}}\text{C}_{\text{2}}\text{a}^{\text{n}+1}\times\text{b}^{2}}{{^\text{n+3}}\text{C}_{\text{3}}\text{a}^{\text{n}}\times\text{b}^{3}}$
Thus, we have
$\frac{\text{T}_{2}}{\text{T}_{3}}=\frac{\text{T}_{3}}{\text{T}_{4}}$
$\Rightarrow \frac{{^\text{n}}\text{C}_{\text{1}}\ \text{a}}{{^\text{n}}\text{C}_{\text{2}}\ \text{b}}=\frac{{^\text{n+3}}\text{C}_{\text{2}}\ \text{a}}{{^\text{n+3}}\text{C}_{\text{3}}\ \text{b}}$
$\Rightarrow \frac{2}{\text{n}-1}=\frac{3}{\text{n}+1}$
$\Rightarrow 2\text{n}+2=3\text{n}-3$
$\Rightarrow \text{n}=5$
View full question & answer→MCQ 441 Mark
The sum of the coefficients of the middle terms of $\left(1+x)^{2 n-1}\right.$ is:
AnswerCorrect option: D. ${ }^{2 n} \mathrm{C}_n$
- ${ }^{2 n} \mathrm{C}_n$
Solution
Consider: $(1+x)^{2 n-1}$
Since $2 \mathrm{n}-1$ is odd, the middle terms are $\left(\frac{2 \mathrm{n}-1+1}{2}\right)^{\text {th }}$ and $\left(\frac{2 \mathrm{n}-1+1}{2}+1\right)^{\text {th }}$ terms
Now, consider the following
$T r+1={ }^n C_r a^{n-r} b^r \ldots \ldots(i)$
Where T represents the term
Here the middle terms are the $n$th term and $(n+1)^{\text {th }}$ term.
$T_{n+1}={ }^{2 n-1} C_n 1^{n-1} x^n={ }^{2 n-1} C_n x^n$
Hence, the coefficient is ${ }^{2 n-1} C_n \ldots$ (1)
$T_n={ }^{2 n-1} C_{n-1} 1^n x^{n-1}={ }^{2 n-1} C_{n-1} x^{n-1}$
Hence, ${ }^{2 n-1} C_{n-1}$ is the coefficient ...(2)
Therefore, sum of the coefficients of the middle terms is
${ }^{2 n-1} C_n+{ }^{2 n-1} C_{n-1}$
$={ }^{2 n} C_n$
View full question & answer→MCQ 451 Mark
Total number of rational terms in the expansion of $(\sqrt[4]{3}+\sqrt[3]{4})^{136}$ is:
AnswerThe number of rational terms is equal to
$\frac{136}{\text{L}.\text{C}.\text{M}(4,3)}+1$
$=\frac{136}{12}+1$
= 11.33 + 1 but we consider the integral value of $\frac{136}{12}.$
=11 + 1
Hence, there are 12 rational terms.
View full question & answer→MCQ 461 Mark
The coefficient of the 8 th term in the expansion of $(1+x)^{10}$ is
- ✓
- B
- C
${ }^{10} \mathrm{C}_8$
- D
Answer
- 120
Solution:
$(1+x)^{10}={ }^{10} \mathrm{C}_0+{ }^{10} \mathrm{C}_1 \mathrm{x}+{ }^{10} \mathrm{C}_2 \mathrm{x}^2+\ldots \ldots . . .+{ }^{10} \mathrm{C}_7 \mathrm{x}^7+{ }^{10} \mathrm{C}_8 \mathrm{x}^8+{ }^{10} \mathrm{C}_9 \mathrm{x}^9+{ }^{10} \mathrm{C}_{10} \mathrm{x}^{10}$
So here, first term is ${ }^{10} \mathrm{C}_0$ then $8^{\text {th }}$ term will be ${ }^{10} \mathrm{C}_7 \mathrm{x}^7$.
$\Rightarrow$ Coefficient of the $8^{\text {th }}$ term $={ }^{10} \mathrm{C}_7$
$=\frac{10!}{7!3!}$
$=\frac{10\times9\times8\times7!}{7!\times3\times2\times1}$
$=120$ View full question & answer→MCQ 471 Mark
If x4 occurs in the rth term in the expansion of $\big(\text{x}^4+\frac{1}{\text{x}^3}\big)15,$ then what is the value of r?
View full question & answer→MCQ 481 Mark
The number of terms with integral coefficient in the expansion of $\Big(17^{\frac{1}{3}}+32^{\frac{1}{2}}\Big)^{300}$ is:
AnswerThe number of rational terms will be
$1+\frac{300}{\text{L}.\text{C}.\text{M}(3,2)}$
$=1+\frac{300}{6}$
$=1+50=51$rational terms.
View full question & answer→MCQ 491 Mark
${ }^{(n+1)} C_1+{ }^{(n+1)} C_2+{ }^{(n+1)} C_3+\ldots . .+{ }^{(n+1)} C_n=$
AnswerCorrect option: B. $2\left(2^n-1\right)$
- $2\left(2^n-1\right)$
Solution
As in the hint required expression $+{ }^{n+1} C_0+{ }^{n+1} C_{n+1}=2^{n+1}$
$\Rightarrow$ required. expression $=2^{n+1}-2=2\left(2^n-1\right)$ View full question & answer→MCQ 501 Mark
Using binomial theorem, the value of $(0.999)^3$ correct to 3 decimal places is:
Answer
- 0.997
Solution:
$(0.999)^3=(1-0.001)^3$
$={ }^3 C_0-{ }^3 C_1(0.001)+{ }^3 C_2(0.001)^2-{ }^3 C_3(0.001)^3$
$=1-0.003+3(0.000001)-(0.000000001)$
$=0.997$ View full question & answer→MCQ 511 Mark
The positive integer just greater than $(1+0.0001)^{10000}$ is:
Answer
- 3
Solution:
$(1+0.0001)^{10000}$
$=\big(1+\frac{1}{10000}\big)^{10000}$
$=\big(1+\frac{1}{\text{n}}\big)^{\text{n}}=\text{n}\text{c}_{0}(1)^{\text{n}}+\text{n}\text{c}_{1}(1)^{\text{n}-1}\cdot\frac{1}{\text{n}}+\text{n}\text{c}_{2}(1)^{\text{n}-2}\frac{1}{\text{n}^{2}}+\ ....$
$=1+\text{n}\cdot\frac{1}{\text{n}}+\frac{\text{n}(\text{n}-1)}{2}\cdot\frac{1}{\text{n}^2}+\ ....$
$=2+\frac{\text{n}(\text{n}-1)}{2\text{n}^{2}}+\ ....$ > 2 Integer just greater than 2 is 3. View full question & answer→MCQ 521 Mark
The total number of rational terms in the expansion of $\Big(7^{\frac{1}{3}}+11^{\frac{1}{9}}\Big)^{6561}$ is:
AnswerTotal number of rational terms will be
$1+\frac{6561}{\text{L}.\text{C}.\text{M}.(3,9)}$
$=1+\frac{6561}{9}$
$=729+1$
$=730$
View full question & answer→MCQ 531 Mark
In the binomial expansion of (a - b)n, n ³ 5 the sum of the 5th and 6th terms is zero. Then $\frac{\text{a}}{\text{b}}$ equals:
- A
$\text{n}-\frac{5}{6}$
- ✓
$\text{n}-\frac{4}{5}$
- C
$\frac{5}{\text{n}-{4}}$
- D
$\frac{6}{\text{n}-{5}}$
AnswerCorrect option: B. $\text{n}-\frac{4}{5}$
View full question & answer→MCQ 541 Mark
In the expansion of $\Big(\text{x}^{2}-\frac{1}{3\text{x}}\Big)^{9},$ the term without x is equal to:
- A
$\frac{28}{81}$
- B
$\frac{-28}{243}$
- ✓
$\frac{28}{243}$
- D
AnswerCorrect option: C. $\frac{28}{243}$
- $\frac{28}{243}$
Solution:
Suppose the $(r + 1)^{th}$ term in the given expansion is independent of x.
Then, we have
$\text{T}_{\text{r}+1}={^\text{9}}\text{C}_{\text{r}}(\text{x}^{2})^{9-\text{r}}\Big(\frac{-1}{3\text{x}}\Big)^{\text{r}}$
$=(-1)^{\text{r}}\ {^\text{9}}\text{C}_{\text{r}}\frac{1}{3\text{r}}\ \text{x}^{18-2\text{r}-\text{r}}$
For this term to be independent of x, we must have
$18-3\text{r}=0$
$\Rightarrow \text{r}=6$
$\therefore$ Required term $=(-1)^{6}\ {^\text{9}}\text{C}_{\text{6}}\ \frac{1}{3^{6}}=\frac{9\times8\times7}{3\times2}\times\frac{1}{3^{6}}=\frac{28}{243}$ View full question & answer→MCQ 551 Mark
If n is an integer lying between 0 and 21, then the least value of n!(21 - n)! is:
Answer
- 11!10!
Solution:
In pascals triangle middle terms has the highest value.
Therefore consider $^{21}C_n$
$=\frac{21!}{(21-\text{n})!\text{n}!}$
For (21 - n)!n!
to be least $^{21}C_n$ has to be maximum.
Therefore, since 21 is odd we have two middle terms $T_{11}$ and $T_{12}$.
Hence for n = 11,
(21 - n)!n! = 11!(10)!
which is less than 12!(9!) for n = 12. View full question & answer→MCQ 561 Mark
The value of $ (126)^{\frac{1}{3}}$ up to three decimal places is:
Answer
- 5.013
Solution:
$ (126)^{\frac{1}{3}}$ can also be written as the cube root of 126.
Hence, $ (126)^{\frac{1}{3}}$is approximately equal to 5.013. View full question & answer→MCQ 571 Mark
The coefficients of $x^p$ and $x^q$ (p and q are positive integers) in the expansion of $(1 + x)^{p+q}$ are:
- ✓
- B
Equal with opposite signs
- C
- D
Answer
- Equal
Solution
The general term is
$t_{r+1}=^{p+q}C_r X^r$
For coefficient of $x^p, r=p$ and hence coefficient is ${ }^{p+q} C_p$
For coefficient of $x^q, r=q$ and hence coefficient is ${ }^{p+q} C_q$
${ }^{p+q} C_p={ }^{p+q} C_q$ View full question & answer→MCQ 581 Mark
The coefficient of $x^4$ in $\Big(\frac{\text{x}}{2}-\frac{3}{\text{x}^{2}}\Big)$ is:
- ✓
$\frac{405}{256}$
- B
$\frac{504}{259}$
- C
$\frac{450}{263}$
- D
AnswerCorrect option: A. $\frac{405}{256}$
- $\frac{405}{256}$
Solution:
Suppose $x^4$ occurs at the $(r+1)^{\text {th }}$ term in the given expansion.
Then, we have
$\text{T}_{\text{r}+1}={^\text{10}}\text{C}_{\text{r}}\Big(\frac{\text{x}}{2}\Big)^{10-\text{r}}\Big(\frac{-3}{2\text{x}^{2}}\Big)$
$=(-1)^{\text{r}}\ {^\text{10}}\text{C}_{\text{r}}\frac{3^{\text{r}}}{2^{10-\text{r}}}\text{x}^{10-\text{r}-2\text{r}}$
For this term to contain $x^4$, we must have
$10-3\text{r}=4$
$\Rightarrow \text{r}=2$
$\therefore$ Required coefficient $={^\text{10}}\text{C}_{\text{2}}\frac{3^{2}}{2^{8}}=\frac{10\times9\times9}{2\times2^{8}}=\frac{405}{256}$ View full question & answer→MCQ 591 Mark
What are the values of k if the term independent of x in the expansion of $\Big(\sqrt{\text{x}}+\frac{\text{k}}{\text{x}^{2}}\Big)^{10}$ is 405?
- ✓
$\pm\ 3$
- B
$\pm\ 6$
- C
$\pm\ 5$
- D
$\pm\ 4$
AnswerCorrect option: A. $\pm\ 3$
View full question & answer→MCQ 601 Mark
The largest coefficient in the expansion of (1 + x)10 is:
AnswerCorrect option: A. $\frac{10!}{(5!)^2}$
- $\frac{10!}{(5!)^2}$
Solution:
Given: (1 + x)10
The greatest coefficient will always occur in the middle term.
Hence, the total number of terms in an expansion is 11. (i.e. 10 + 1 = 11)
Therefore, middle term $ =\Big[\big(\frac{10}{2}\big)+1\Big]=5+1=6\text{th }\text{term}.$
So, $\mathrm{T} 6={ }^{10} \mathrm{C}_5 \times \mathrm{x}^5$
Therefore, the coefficient of the greatest term $={ }^{10} \mathrm{C}_5=\frac{10!}{(5!)^2}$. View full question & answer→MCQ 611 Mark
The coefficient of $x^3y^4$ in $(2x + 3y^2)^5$ is:
Answer
- 720
Solution:
Given: $\left(2 x+3 y^2\right)^5$
Therefore, the general form for the expression $\left(2 x+3 y^2\right)^5$ is $T_{r+1}={ }^5 C_{r \times}(2 x)^r \times\left(3 y^2\right)^{5-r}$
Hence, $T_{3+1}={ }^5 C_3(2 x)^3 \times\left(3 y^2\right)^{5-3}$
$T_4={ }^5 C_3(2 x)^3 \times\left(3 y^2\right)^2$
$T 4={ }^5 C_3 \times 8 x^3 \times 9 y^4$
On simplification, we get
$T_4=720 x^3 y^4$
Therefore, the coefficient of $x^3 y^4$ in $\left(2 x+3 y^2\right)^2$ is 720 . View full question & answer→MCQ 621 Mark
If A and B are the sums of odd and even terms respectively in the expansion of $(\text{x}+\text{a})^{\text{n}},$ then $(\text{x}+\text{a})^{\text{2n}}-(\text{x}-\text{a})^{2\text{n}}$ is equal to:
AnswerIf A and B denote respectively the sums of odd terms and even terms in the expansion $(\text{x}+\text{a})^{\text{n}}$
Then, $(\text{x}+\text{a})^{\text{n}}=\text{A}+\text{B}\ ...(\text{i})$
$(\text{x}-\text{a})^{\text{n}}=\text{A}-\text{B}\ ...(\text{ii})$
Squaring and subtraction equation (ii) from(i) we get,
$ (\text{x}+\text{a})^{2\text{n}}-(\text{x}-\text{a})^{2\text{n}}=(\text{A}+\text{B})^{2}-(\text{A}-\text{B})^{2}$
$\Rightarrow (\text{x}+\text{a})^{2\text{n}}-(\text{x}-\text{a})^{2\text{n}}=4\text{AB}$
View full question & answer→MCQ 631 Mark
If the fifth term of the expansion $\Big(\text{a}^{\frac{2}{3}}+\text{a}^{-1}\Big)^{\text{n}}$ does not contain 'a'. Then n is equal to:
Answer$\text{T}_{5}=\text{T}_{4+1}$
$={^\text{n}}\text{C}_{\text{4}}\big(\text{a}^{\frac{2}{3}}\big)^{\text{n-4}}(\text{a}^{-1})^{4}$
$={^\text{n}}\text{C}_{\text{4}}\ \text{a}^{\big(\frac{2\text{n}-8}{3}-4\big)}$
For this term to be independent of a, we must have
$\frac{2\text{n}-8}{3}-4=0$
$\Rightarrow 2\text{n}-20=0$
$\Rightarrow \text{n}=10$
View full question & answer→MCQ 641 Mark
Choose the correct answer.The two successive terms in the expansion of $(1 + x)^{24}$ whose coefficients are in the ratio 1 : 4 are:
$[\text{Hint}:\frac{^{24}\text{C}_\text{r}}{^{24}\text{C}_{\text{r}+1}}=\frac{1}{4}\ \frac{\text{r}+1}{24-\text{r}}\ \frac{1}{4}\Rightarrow4\text{r}+4=24-4\Rightarrow\text{r}=4]$
- A
$3^{\text {rd }}$ and $4^{\text {th }}$.
- B
$4^{\text {th }}$ and $5^{\text {th }}$.
- ✓
$5^{\text {th }}$ and $6^{\text {th }}$.
- D
$6^{\text {th }}$ and $7^{\text {th }}$.
AnswerCorrect option: C. $5^{\text {th }}$ and $6^{\text {th }}$.
- $5^{\text {th }}$ and $6^{\text {th }}$.
Solution:
Let the two successive terms in the expansion of $(1 + x)^{24}$ be $(r + 1)(r + 2)^{th}$ terms.
Now, $\text{T}_{\text{r}+1}=\ ^{24}\text{C}_\text{r}\text{x}^\text{r}\ \text{and}\ \text{T}_{\text{r}+2}=\ ^{24}\text{C}_{\text{r}+1}\text{x}^{\text{r}+1}$
Given that, $\frac{^{24}\text{C}_\text{r}}{^{24}\text{C}_{\text{r}+1}}=\frac{1}{4}$
$\Rightarrow\frac{\frac{(24)!}{\text{r}!(24-\text{r})!}}{\frac{(24)!}{(\text{r}+1)!(24-\text{r}-1)!}}=\frac{1}{4}$ $\Rightarrow\frac{(\text{r}+1)\text{r}!(23-\text{r})!}{\text{r}!(24-\text{r})(23-\text{r})!}=\frac{1}{4}\Rightarrow\frac{\text{r}+1}{24-\text{r}}=\frac{1}{4}$
$\Rightarrow4\text{r}+4=24-\text{r}\Rightarrow\text{r}=4$
$\therefore\text{T}_{4+1}=\text{T}_5\ \text{and}\ \text{T}_{4+2}=\text{T}_6$
Hence. $5^{th}$ and $6^{th}$ terms. View full question & answer→MCQ 651 Mark
The number of terms with integral coefficient in the expansion of $\Big((27)^{\frac{1}{6}}+\sqrt[10]{32\text{x}}\Big)^{600}$ is:
Answer$\Big(27^{\frac{1}{6}}+32^{\frac{1}{10}}\text{x}\Big)^{600}$
$=\Big(3^{\frac{1}{2}}+2^{\frac{1}{2}}\Big)^{600}$
Total number of integral terms will be
$=\frac{600}{\text{L}.\text{C}.\text{M}(2,2)}+1$
$=\frac{600}{2}+1$
$=301$
View full question & answer→MCQ 661 Mark
The total number of terms in the expansion of (x + a)51 - (x - a)51 after simplification is:
MCQ 671 Mark
The number of rational terms in the expansion of $(9^{\frac{1}{4}}+8^{\frac{1}{6}})^{1000}$ is:
AnswerThe general term in the expansion of $(9^{\frac{1}{4}}+8^{\frac{1}{6}})^{1000}$ is
$\text{T}_{\text{r}+1}=^{1000}\text{C}\text{r}\big(9^{\frac{1}{4}}\big)^{1000-\text{r}}+\big(8^{\frac{1}{6}}\big)^{\text{r}}$
$=^{1000}\text{C}_{\text{r}}3\frac{1000-\text{r}}{2}2\frac{1}{2}$
The above term will be rational if exponent of 3 and 2 are integers.
It means $\frac{1000-\text{r}}{2}$ and $\frac{\text{r}}{2}$ must be integers
The possible set of values of r is {0, 2, 4, ...., 1000}
Hence, number of rational terms is 501.
View full question & answer→MCQ 681 Mark
The total number of terms in the expansion of $(\text{x}+\text{a})^{100}+(\text{x}-\text{a})^{100}$ after simplification is:
AnswerHere, n i.e. 100 is even.
$\therefore$ Total number of terms in the expansion $=\frac{\text{n}}{2}+1=\frac{100}{2}+1=51$
View full question & answer→MCQ 691 Mark
The total number of terms which are dependent on the value of x in the expansion of $\Big(\text{x}^2-2+\frac{1}{\text{x}^{2}}\Big)^{\text{n}}$ is equal to:
Answer$\Big(\text{x}^2-2+\frac{1}{\text{x}^{2}}\Big)^{\text{n}}$
$=\Big(\big(\text{x}-\frac{1}{\text{x}}\big)^{2}\Big)^{\text{n}}$
$=\big(\text{x}-\frac{1}{\text{x}}\big)^{{2}{\text{n}}}$
$\text{T}_{\text{r}+1}={^{10}}\text{C}_{\text{r}}\text{x}^{2\text{n}-2\text{r}}$
For term independent of x
2n - 2r = 0
n = r
Hence there will be 1 term independent of x.
Since the total number of terms are 2n + 1.
Hence the total number of term dependent on x will be
Total number of terms - (total number of terms independent of x).
= 2n + 1 − 1
= 2n
View full question & answer→MCQ 701 Mark
If the 4th term in the binomial expansion of (p + 1)n is $\frac{5}{2}$ then:
- A
$\text{n}=8, \text{p}=6$
- B
$\text{n}=8, \text{p}=\frac{1}{2}$
- ✓
$\text{n}=6, \text{p}=\frac{1}{2}$
- D
$\text{n}=6, \text{p}=6$
AnswerCorrect option: C. $\text{n}=6, \text{p}=\frac{1}{2}$
- $\text{n}=6, \text{p}=\frac{1}{2}$
View full question & answer→MCQ 711 Mark
Find the sum of coefficient of middle terms of the expansion $\Big(3\text{x}-\frac{\text{x}^3}{6}\Big)^7:$
AnswerCorrect option: A. $\frac{595}{48}$
- $\frac{595}{48}$
Solution
Total number of terms are 8.
So, middle term will be the $4^{th}$ and $5^{th}$ term.
$\therefore\text{t}_{3+1}=^7\text{C}_3(-1)^{3}(3\text{x})^{7-3}\big(\frac{\text{x}^3}{6}\big)^3=-\frac{105\text{x}^{13}}{8}$
$\therefore\text{t}_{4+1}=^7\text{C}_4(-1)^{4}(3\text{x})^{7-4}\big(\frac{\text{x}^3}{6}\big)^4=\frac{35\text{x}^{15}}{48}$
So, $-\frac{105}{8}+\frac{35}{48}=-\frac{595}{48}$ View full question & answer→MCQ 721 Mark
The middle term in the expansion of $\Big(\frac{2\text{x}^{2}}{3}+\frac{3}{2\text{x}^{2}}\Big)^{10}$ is:
Answer
- 252
Solution:
Hence, n, i.e., 10, is an even number.
$\therefore$ Middle term $=\Big(\frac{10}{2}+1\Big)^{\text{th}}$ term = $6^{th}$ term
Thus, we have
$\text{T}_{6}=\text{T}_{5+1}$
$={^\text{10}}\text{C}_{\text{5}}\Big(\frac{2\text{x}^{2}}{3}\Big)^{10-5}\Big(\frac{3}{2\text{x}^{2}}\Big)^{5}$
$=\frac{10\times9\times8\times7\times6}{5\times4\times3\times2}\times\frac{2^{5}}{3^{5}}\times\frac{3^{5}}{2^{5}}$
$=252$ View full question & answer→MCQ 731 Mark
If $\text{r}^{th}$ term is the middle term in the expansion of $\Big(\text{x}^{2}-\frac{1}{2\text{x}}\Big)^{20},$ then $(r+3)^{\text {th }}$ term is:
- A
${^\text{20}}\text{C}_{\text{14}}\ \Big(\frac{\text{x}}{2^{14}}\Big)$
- B
${^\text{20}}\text{C}_{\text{12}}\ \text{x}^{2}\ 2^{-12}$
- ✓
$-{^\text{20}}\text{C}_{\text{7}}\ \text{x}\ 2^{-13}$
- D
AnswerCorrect option: C. $-{^\text{20}}\text{C}_{\text{7}}\ \text{x}\ 2^{-13}$
- $-{^\text{20}}\text{C}_{\text{7}}\ \text{x}\ 2^{-13}$
Solution:
Here, n is even,
So, The middle term in the given expansion is $\Big(\frac{20}{2}+1\Big)^{\text{th}}=11^{\text{th}}$
Therefore, $(r+3)^{\text {th }}$ term is the $14^{\text {th }}$ term
$\text{T}_{14}={^\text{20}}\text{C}_{\text{13}}(\text{x}^{2})^{20-13}\ \Big(\frac{-1}{2\text{x}}\Big)$
$=(-1)^{13}\ {^\text{20}}\text{C}_{\text{13}}\ \frac{\text{x}^{14-3}}{2^{13}}$
$=-{^\text{20}}\text{C}_{\text{7}}\ \text{x}\ 2^{-13}$ View full question & answer→MCQ 741 Mark
If in the expansion of $\Big(\text{x}^{4}-\frac{1}{\text{x}^{3}}\Big)^{15},\text{x}^{-17}$ occurs in rth term, then
Answer
- r = 12
Solution:
Here,
$\text{T}^{\text{r}}={^\text{15}}\text{C}_{\text{r}-1}(\text{x}^{4})^{15-\text{r}+1}\Big(\frac{-1}{\text{x}^{3}}\Big)^{\text{r}-1}$
$=(-1)^{\text{r}}\times{^\text{15}}\text{C}_{\text{r}-1}\ \text{x}^{64-4\text{r}-3\text{r}+3}$
For this term to contain $x^{-17}$, we must have
$67-7\text{r}=-17$
$\Rightarrow \text{r}=12 $ View full question & answer→MCQ 751 Mark
In the expansion of $\big(\frac{\text{x}+2}{\text{x}^2}\big)15,$ the term independent of x is:
View full question & answer→MCQ 761 Mark
${ }^{15} C_3+{ }^{15} C_5+\ldots .+{ }^{15} C_{15}$ will be equal to:
- A
$2^{14}$
- ✓
$2^{14}-15$
- C
$2^{14}+15$
- D
$2^{14}-1$
AnswerCorrect option: B. $2^{14}-15$
- $2^{14}-15$
Solution
We know
${ }^{15} C_1+{ }^{15} C_3+{ }^{15} C_5+\ldots . .{ }^{15} C_{15}=2^{15-1}$
$\therefore{ }^{15} \mathrm{C} 3+{ }^{15} C_5+\ldots+{ }^{15} C_{15}=2^{14}-15$ View full question & answer→MCQ 771 Mark
The coefficients of the expansions are arranged in an array. This array is called ………
MCQ 781 Mark
The number of terms in the expansion of $[(a + 4b)^3(a - 4b)^3]^2$ are:
Answer
- 7
Solution:
$[(a + 4b)^3 (a - 4b)^3]^2$
$= [(a + 4b)(a - 4b)]^6$
$= [a^2 - 16b^2]^6$
Hence total number of terms is n + 1
Here n = 6
Therefore, total number of terms is 7. View full question & answer→MCQ 791 Mark
Expand the following binomials: $(x-3)^5$
- A
$x^5+25 x^4+90 x^3-270 x^2+405 x-243$
- B
$x^5-15 x^4+90 x^3-270 x^2-405 x-243$
- C
$x^5-15 x^4+80 x^3-270 x^2+405 x-243$
- ✓
$x^5-15 x^4+90 x^3-270 x^2+405 x-243$
AnswerCorrect option: D. $x^5-15 x^4+90 x^3-270 x^2+405 x-243$
- $x^5-15 x^4+90 x^3-270 x^2+405 x-243$
Solution:
$(x-3)^5={ }^5 C_0 x^5+{ }^5 C_1 x^4(-3)^1+{ }^5 C_2 x^3(-3)^2+{ }^5 C_3 x^2(-3)^3+{ }^5 C_4 x(-3)^4+{ }^5 C_5(-3)^5$
$=x^5-15 x^4+90 x^3-270 x^2+405 x-243$ View full question & answer→MCQ 801 Mark
${ }^{(2 n+1)} C_0{ }^{-(2 n+1)} C_1+{ }^{(2 n+1)} C_2-\ldots .{ }^{2 n+1} C_{2 n}=$
Answer
- 1
Solution:
In some questions, substituting n = a positive number in both the question and the answer is the fastest way to achieve the correct option.
Although there is always alternate options like writing the general term and splitting it to a format that can be solved but that takes long in a limited time paper.
Try to put n = 1.
In the end only option A remains.
View full question & answer→MCQ 811 Mark
The term independent of x in the expansion of $\Big(2\text{x}+\frac{1}{3\times2}\Big)^9.$
View full question & answer→MCQ 821 Mark
The number of terms whose values depends on x in the expansion of $\Big(\text{x}^{2}-2+\frac{1}{\text{x}^{2}}\Big)^{\text{n}}$ is:
Answer
- 2n
Solution:
$\Big(\text{x}^{2}-2+\frac{1}{\text{x}^{2}}\Big)^{\text{n}}$
$=\Big[\big(\text{x}-\frac{1}{\text{x}}\big)^{2}\Big]^{\text{n}}$
$=\big(\text{x}-\frac{1}{\text{x}}\big)^{{2}\text{n}}$
Hence there will be $2 \mathrm{n}+1$ terms.
The middle term i.e $\mathrm{n}+1^{\text {th }}$ term will be independent of x .
Hence total number of terms, dependent on x will be
$2 n+1-(1)$
$=2 n \text { terms. }$ View full question & answer→MCQ 831 Mark
The sum of the coefficients of the first 10 terms in the expansion of $(1-x)^{-3}$
Answer
- 220
Solution
For $(1-\mathrm{x})^{-3}$, the sum of the first r terms will be
${ }^{n+r-1} C_n$
Replacing $\mathrm{n}=3$ and $\mathrm{r}=10$ in the above formula, we get
$3+10-1 C_3$
$={ }^{12} C_3$
$=\frac{12.11 \cdot 10}{3!}$
$=\frac{12.11 \cdot 10}{6}$
$=2.11 .10$
$=220$ View full question & answer→MCQ 841 Mark
The coefficient of $x^3$ in $\Big(\sqrt{\text{x}^5}+\frac{3}{\sqrt{\text{x}^{3}}}\Big)^5$ is:
Answer
- 540
Solution:
$\text{r}=\frac{6\times\frac{5}{2}-3}{\frac{5}{2}+\frac{3}{2}}=\frac{15-3}{4}=3$
$\therefore$ Coefficient of $x^3 is ^6C_33^3$
$=\frac{6\times5\times4}{3\times2\times1}\cdot27$
= 5 × 4 × 27 = 540 View full question & answer→MCQ 851 Mark
What is the approximate value of (1.02)8?
MCQ 861 Mark
The sum of the coefficients in the expansion of $(1 + 5x - 7x^3)^{3165}$ is:
Answer
- -1
Solution:
To get sum of coefficient put x = 1
Hence sum of the coefficients in the expansion of $(1 + 5x - 7x^3)^{3165}$ is,
$= (1 + 5 - 7)^{3165} = (-1)⋅(−1)^{3164} = -1$ View full question & answer→MCQ 871 Mark
Number of rational term is the expansion of $(7^{\frac{1}{3}}+1^{\frac{11}{9}})^{729}$
AnswerSince 7 and 11 are prime numbers, hence application of general formula for number of rational terms will be
$=1+\frac{729}{\text{L}.\text{C}.\text{M}(1,3)}$
$=1+\frac{729}{9}$
$=1+81$
$=82$ rational terms.
View full question & answer→MCQ 881 Mark
The coefficient of $x^5$ in the expansion of $\left(1+x_2\right)^5(1+x)^4$ is?
Answer
- Zero
Solution:
according to binomial expansion,
$\left(1+x^2\right)^5=5 C_0 \times 1+5 C_1\left(x^2\right)+5 C_2\left(x^2\right)^2+5 C_3\left(x^2\right)^3+5 C_4\left(x^2\right)^4+5 C_5\left(x_2\right)^5$
$=1+5 x^2+10 x^4+10 x^6+5 x^8+x^0 \Longrightarrow(1)$
$=(1+x)^4=1+4 C_1 x+4 C_2 x^2+4 C_3 x^3+4 C_4 x^4 \Rightarrow(2)$
from 1 and 2 we can see that coefficient of 5 is not there so $x^5$ is zero. View full question & answer→MCQ 891 Mark
If the coefficient of (2r + 4)th term and $(r – 2)^{th}$ term in the expansion of $(1 + x)^{18}$ are equal, then r is equal to:
View full question & answer→MCQ 901 Mark
In the expansion of $\Big(\text{x}-\frac{1}{\text{x}}\Big)^6,$ the constant term is:
Answer
- -20
Solution:
In expansion of $\Big(\text{x}-\frac{1}{\text{x}}\Big)^6,$ the constant term is ${^{6}}\text{C}_{\text{k}}\text{x}^{6-\text{k}}\big(\frac{-1}{\text{x}}\big)^{\text{k}}$
In constant term, power of x must be zero
$\therefore$ 6 - k - k = 02k = 6.k = 3
Hence, the constant term is $- ^6c_3 = -20$. View full question & answer→MCQ 911 Mark
If $\boldsymbol{n}$ is a positive integer, then the number of terms in the expansion of $(x+a)^n$ is:
Answer
- n + 1
Solution:
In binomial expansion the terms goes from $\mathrm{nC}_0 x^n$ to $\mathrm{nC}_n \mathrm{a}^n$ i.e the base of C goes from 0 to n and this shows that there must be $(n+1)$ terms. View full question & answer→MCQ 921 Mark
The sum of the coefficients in the expansion of $(x+2 y+z)^{10}$ is:
- A
$2^{10}$
- ✓
$4^{10}$
- C
$3^{10}$
- D
AnswerCorrect option: B. $4^{10}$
- $4^{10}$
Solution:
Given expression is $(x+2 y+z)^{10}$ Substituting $x=y=z=1$, we get the sum of the coefficients as
$(1+2+1)^{10}$
$=4^{10}$. View full question & answer→MCQ 931 Mark
Choose the correct answer.If the coefficients of $2^{\text {nd }}, 3$ rd and the $4^{\text {th }}$ terms in the expansion of $(1+x)^n$ are in A.P., then value of $n$ is:
[Hint: $2^n C_2={ }^n C_1+{ }^n C_3 \Rightarrow n^2-9 n+14=0 \Rightarrow n=2$ or 7.$]$
Answer
- 7.
Solution:
$(1+\text{x})^\text{n}=\ ^\text{n}\text{C}_0+\ ^\text{n}\text{C}_1\text{x}+\ ^\text{n}\text{C}_2\text{x}^2+\ ^\text{n}\text{C}_3\text{x}^3+...+\ ^\text{n}\text{C}_\text{n}\text{x}^\text{n}$
So, coefficients of $2^{nd}, 3^{rd}$ and 4th terms are $^nC_1, ^nC_2$ and $^nC_3$, respectively.
Given that, $^nC_1, ^nC_2$ and $^nC_3$, are in A.P.
$\therefore2\ ^\text{n}\text{C}_2=\ ^\text{n}\text{C}_1+\ ^\text{n}\text{C}_3$
$\Rightarrow2\Big[\frac{\text{n}!}{(\text{n}-2)!2!}\Big]=\text{n}+\frac{\text{n}!}{3!(\text{n}-3)!}$
$\Rightarrow2\Big[\frac{\text{n}(\text{n}-1)}{2!}\Big]=\text{n}+\frac{\text{n}(\text{n}-1)(\text{n}-2)}{3!}\Rightarrow(\text{n}-1)=1+\frac{(\text{n}-1)(\text{n}-2)}{6}$
$\Rightarrow6\text{n}-6=6+\text{n}^2-3\text{n}+2\Rightarrow\text{n}^2-9\text{n}+14=0$ $\Rightarrow(\text{n}-7)(\text{n}-2)=0$
$\therefore\text{n}=2\ \text{or}\ \text{n}=7$
Since n = 2 is not possible, so n = 7. View full question & answer→MCQ 941 Mark
The coefficient of $x^8 y^{10}$ in the expansion of $(x+y)^{18}$ is:
AnswerCorrect option: A. ${^\text{18}}\text{C}_{\text{8}}$
- ${^\text{18}}\text{C}_{\text{8}}$
Solution:
Suppose $(r+1)^{\text {th }}$ term in the given expansion is independent of x.
Then, we have
$\text{T}_{\text{r}+1}={^\text{18}}\text{C}_{\text{r}}(\text{x})^{18-\text{r}}\ \text{y}^{\text{r}}$
For this term to be independent of x, we must have
$\text{r}=10$
Hence, the required coefficient is ${^\text{18}}\text{C}_{\text{10}}$ or ${^\text{18}}\text{C}_{\text{8}}$ View full question & answer→MCQ 951 Mark
If $ \text{z}=\Big(\frac{\sqrt{3}}{2}+\frac{\text{i}}{2}\Big)5+\Big(\frac{\sqrt{3}}{2}-\frac{\text{i}}{2}\Big)5,$ then:
View full question & answer→MCQ 961 Mark
The number of terms in the expression of $(x+y)^{n-1}$ is 2018 then n.
Answer
- 2018
Solution:
Here, the no. of terms in binomial expansion of $(x+y)^{n-1}$ is $(n-1)+1$ i.e; one more the exponent,
$\Rightarrow(n-1)+1=2018$
$\Rightarrow n=2018$ View full question & answer→MCQ 971 Mark
The coefficient of $x^{18}$ in the product $(1+x)(1-x)^{10}\left(1+x+x^2\right)^9$ is?
Answer
- 84
Solution:
$(1+x)(1-x)^{10}\left(1+x+x^2\right)^9$
$\left(1-x^2\right)\left(1-x^3\right)^9$
${ }^9 C_6=84$. View full question & answer→MCQ 981 Mark
If coefficient of $x^{100}$ in $1+(1+x)(1+x)^2+\ldots . .+(1+x)^n$ (if $n \geq 100$ ) is $C_{101}^{201}$ then the value of $n$ equals.
Answer
- 200
Solution:
${ }^n C_r+{ }^n C_{(r+1)}={ }^{(n+1)} C_{(r+1)}$
coefficient of $x^{100}$ is ${ }^{100} \mathrm{C}_{100}+{ }^{101} \mathrm{C}_{100}+{ }^{102} \mathrm{C}_{100}+\ldots . . . .+{ }^n \mathrm{C}_{100}$.
Which is equal to ${ }^{(n+1)} \mathrm{C}_{101}$.
Therefore, $\mathrm{n}+1=201$
Which implies $n=200$ View full question & answer→MCQ 991 Mark
The middle term in the expansion of $\Big(\frac{2\text{x}}{3}=\frac{3}{2\text{x}^{2}}\Big)^{2\text{n}}$ is:
- A
${^\text{2n}}\text{C}_{\text{n}}$
- ✓
$(-1)\ {^\text{2n}}\text{C}_{\text{n}}\ \text{x}^{-\text{n}}$
- C
${^\text{2n}}\text{C}_{\text{n}}\ \text{x}^{-\text{n}}$
- D
AnswerCorrect option: B. $(-1)\ {^\text{2n}}\text{C}_{\text{n}}\ \text{x}^{-\text{n}}$
Here, n is even,
Middle term in the given expansion $=\Big(\frac{2\text{n}}{2}+1\Big)^{\text{th}}=(\text{n}+1)$
$={^\text{2n}}\text{C}_{\text{n}}\Big(\frac{2\text{x}}{3}\Big)^{2\text{n}-\text{n}}\ \Big(\frac{-3}{2\text{x}^{2}}\Big)^{\text{n}}$
$(-1)\ {^\text{2n}}\text{C}_{\text{n}}\ \text{x}^{-\text{n}}$
View full question & answer→MCQ 1001 Mark
The coefficient of the term independent of x in the expansion of $\Big(\text{ax}+\frac{\text{b}}{\text{x}}\Big)^{14}$ is:
- A
$14!\ \text{a}^{7}\ \text{b}^{7}$
- B
$\frac{14!}{7!}\ \text{a}^{7}\ \text{b}^{7}$
- ✓
$\frac{14!}{(7!)^{2}}\ \text{a}^{7}\ \text{b}^{7}$
- D
$\frac{14!}{(7!)^{3}}\ \text{a}^{7}\ \text{b}^{7}$
AnswerCorrect option: C. $\frac{14!}{(7!)^{2}}\ \text{a}^{7}\ \text{b}^{7}$
- $\frac{14!}{(7!)^{2}}\ \text{a}^{7}\ \text{b}^{7}$
Solution:
Suppose $(r + 1)^{th}$ term in the given expansion is independent of x.
Then, we have
$\text{T}_{\text{r}+1}={^\text{14}}\text{C}_{\text{r}}(\text{ax})^{14-\text{x}}\ \Big(\frac{\text{b}}{\text{a}}\Big)^{\text{r}}$
$={^\text{14}}\text{C}_{\text{r}}\ \text{a}^{14-\text{r}}\ \text{b}^{\text{r}}\ \text{x}^{14-2\text{r}}$
For this term to be independent of x, we must have
$=14-2\text{r}=0$
$\Rightarrow \text{r}=7$
Required term $= {^\text{14}}\text{C}_{\text{7}}\ \text{a}^{14-7}\ \text{b}^{7}=\frac{14!}{(7)!}\ \text{a}^{7}\ \text{b}^{7}$ View full question & answer→MCQ 1011 Mark
What is the coefficient of $x^3 y^4$ in $\left(2 x+3 y^2\right) 5$ ?
View full question & answer→MCQ 1021 Mark
The value of $(\sqrt{5}+1)^{5}-(\sqrt{5}-1)^{5} $ is:
Answer
- 352
Solution:
Given, $(\sqrt{5}+1)^{5}-(\sqrt{5}-1)^{5} $
$=(\sqrt{5}+1)^{5}-(\sqrt{5}-1)^{5} $
The even terms will get eliminated.
Hence, we get
$2\left[{ }^5 \mathrm{C}_0+{ }^5 \mathrm{C}_2 5+{ }^5 \mathrm{C}_4 5^2\right]$
$=2[1+50+125]$
$=2[176]$
$=352 .$ View full question & answer→MCQ 1031 Mark
In any binomial expansion, the number of terms are:
- A
$\geq5$
- ✓
$\geq2$
- C
$\geq3$
- D
$\geq4$
AnswerCorrect option: B. $\geq2$
Bi - nomial, involves summation of two terms.
Let the terms be x and y.
Therefore a binomial expansion can be of the form, (x + y)n.
where $\text{n}\geq1$ If n = 1, we get only two terms.If n > 1 where n is an integer, then it gives us in total (n + 1) terms.
Thus, number of terms has to be $\geq2.$
View full question & answer→MCQ 1041 Mark
The number of real negative terms in the binomial expansion of $(1+i x)^{4 n-2}$, $\text{n}\in\text{N},$ x > 0, is:
Answer
- n
Solution:
$(1+i x)^{4 n-2}$
$=\left((1+i x)^2\right)^{2 n-1}$
$=\left(1-x^2+2 i x\right)^{2 n-1}$
$=\left[\left(1-x^2\right)+i(2 x)\right]^{2 n-1}$
Total number of terms will be $2 n-1+1=2 n$.
Hence the number of real negative terms will therefore be $=\frac{2 \mathrm{n}}{2}=\mathrm{n}$. View full question & answer→MCQ 1051 Mark
Sum of the coefficients of the terms of degree $m m$ in the expansion of $(1+x)^n(1+y)^n(1+z)^n$ is:
- A
$\left({ }^n \mathrm{C}_{\mathrm{m}}\right)^3$
- B
$3\left({ }^n \mathrm{C}\right.$ m)
- C
${ }^n \mathrm{C}_{3 \mathrm{~m}}$
- ✓
${ }^{3 n} \mathrm{C}_{\mathrm{m}}$
AnswerCorrect option: D. ${ }^{3 n} \mathrm{C}_{\mathrm{m}}$
- ${ }^{3 n} \mathrm{C}_{\mathrm{m}}$
Solution:
The Coefficient of $x^m=$ Number of ways of choosing $m$ balls out of n black balls, n green balls and n blue ball.
Hence total number of balls $=3 \mathrm{n}$.
Required is m .
Hence required combination is ${ }^{3 n} \mathrm{C}_{\mathrm{m}}$.
Hence the coefficient of $x^m$ in $(1+x)^n(1+y)^n(1+z)^n$
$={ }^{3 n} C_m.$ View full question & answer→MCQ 1061 Mark
If the coefficients of the $(n + 1)^{th}$ term and the $(n + 3)^{th}$ term in the expansion of $(1+\text{x})^{20}$ are equal, then the value of n is:
Answer
- 9
Solution:
Coefficient of $(r + 1)^{th}$ term = Coefficient of $(n + 3)^{th}$
Then, we have
${^\text{20}}\text{C}_{\text{n}}={^\text{20}}\text{C}_{\text{n}+2}$
$\Rightarrow 2\text{n}+2=20$
$\Rightarrow \text{n}=9$ View full question & answer→MCQ 1071 Mark
In the expansion of $(1+x)^n$, the sum of coefficients of odd powers of $x$ is:
- A
$2^n+1$
- B
$2^n-1$
- C
$2^n$
- ✓
$2^{n-1}$
AnswerCorrect option: D. $2^{n-1}$
- $2^{n-1}$
Solution:
$(1+x)^n=C_0+C_{1 x}+C_2 x^2+C_3 x^3+\ldots+C_n x^n$
Putting $x=1$ and $x=1$ and subtracting, we get.
$2^n=2\left(C_1+C_3+C_5+\ldots\right)$
$\therefore C_1+C_3+C_5+\ldots=2^{n-1}$
Or the sum of the coefficients of the odd power of $x$ is $2^{n-1}$. View full question & answer→MCQ 1081 Mark
If an the expansion of $(1+\text{x})^{15},$ the coefficients of $(2\text{r}+3)^{\text{th}}$ and $(\text{r}-1)^{\text{th}}$ terms are equal, then the value of r is:
AnswerCoefficients of $(2\text{r}+3)^{\text{th}}$ and $(\text{r}-1)^{\text{th}}$ terms in the given expansion are ${^\text{15}}\text{C}_{\text{2r}+2}$ and ${^\text{15}}\text{C}_{\text{2r}-2}.$
Then, we have
${^\text{15}}\text{C}_{\text{2r}+2}={^\text{15}}\text{C}_{\text{r}-2}$
$\Rightarrow 2\text{r}+2=\text{r}-2$ or $2\text{r}+2+\text{r}-2=15$
$\Rightarrow \text{r}=-4$ or $\text{r}=5$
Neglecting the negative value, We have
$\text{r}=5$
View full question & answer→MCQ 1091 Mark
If the coefficients of $x^7$ and $x^8$ in $\left(2+\frac{x}{3}\right) n$ are equal, then $n$ is:
View full question & answer→MCQ 1101 Mark
The coefficient of y in the expansion of $\Big(\text{y}^2+\big(\frac{\text{c}}{\text{y}}\big)\Big)^5$ is:
- A
$10c$
- B
$29c$
- C
$10 c^3$
- ✓
$20 c^3$
AnswerCorrect option: D. $20 c^3$
- $10 c^3$
Solution:
Given:$\Big(\text{y}^2+\big(\frac{\text{c}}{\text{y}}\big)\Big)^5$
$\Big(\text{y}^2+\big(\frac{\text{c}}{\text{y}}\big)\Big)^{5}=\ ^{5}\text{C}_{\text{r}}\times(\text{y}^2)^{\text{r}}\times\big(\frac{\text{c}}{\text{y}}\big)^{5-\text{r}}$
$\Big(\text{y}^2+\big(\frac{\text{c}}{\text{y}}\big)\Big)^{5}={5}\text{C}{\text{r}}\times\text{y}^{2{\text{r}}}\times\big(\frac{\text{c}^{5-\text{r}}}{\text{y}^{5-\text{r}}}\big)$
On solving this, we get $r=3$.
Hence, the coefficient of $y={ }^5 \mathrm{C}_3 \times \mathrm{c}^3=10 \mathrm{c}^3$. View full question & answer→MCQ 1111 Mark
If the coefficient of x in $\Big(\text{x}^{2}+\frac{\lambda}{\text{x}}\Big)^{5}$ is 270, then $\lambda=$
Answer
- 3
Solution:
The coefficient of x in the given expansion where x occurs at the $(r + 1)^{th}$ term.
We have,
${^\text{15}}\text{C}_{\text{r}}(\text{x}^{2})^{5-\text{r}}\ \Big(\frac{\lambda}{\text{x}}\Big)^{\text{r}}$
$={^\text{15}}\text{C}_{\text{r}}\ \lambda^{\text{r}}\ \text{x}^{10-2\text{r}-\text{r}}$
For it to contain x, we must have
$10-3\text{r}=1$
$\Rightarrow \text{r}=3$
Coefficient of x in the given expansion,
$={^\text{15}}\text{C}_{\text{3}}\ \lambda^{\text{3}}=10\lambda^{3}$
Now, we have
$10\lambda^{3}=270$
$\Rightarrow \lambda^{3}=27$
$\Rightarrow \lambda=3$ View full question & answer→MCQ 1121 Mark
Number of rational terms in the expansion of $\Big(\sqrt{2}+\sqrt[4]{3}\Big)^{100}$ is:
Answer
- 26
Solution:
The general term for the following expression is $\text{T}_{\text{r}+1}={^{100}}\text{C}_{\text{r}}2^{50-\frac{\text{r}}{2}}\cdot3\frac{\text{r}}{4}.$
Hence we get rational terms for
$r = 0, 4, 8, 12 ....100$
$a_n = a + (n - 1).d$
$100 = 0 + (n - 1).4$
$25 = n - 1$
$n = 26$ View full question & answer→MCQ 1131 Mark
The coefficient of $\frac{1}{\text{x}}$ in the expansion of $(1+\text{x})^{\text{n}}+\Big(1+\frac{1}{\text{x}}\Big)^{\text{n}}$ is:
- A
$\frac{\text{n}!}{\big[(\text{n}-1)!(\text{n}+1)!\big]}$
- ✓
$\frac{(\text{2n})!}{\big[(\text{n}-1)!(\text{n}+1)!\big]}$
- C
$\frac{(\text{2n})!}{\big[(\text{2n}-1)!(\text{2n}+1)!\big]}$
- D
AnswerCorrect option: B. $\frac{(\text{2n})!}{\big[(\text{n}-1)!(\text{n}+1)!\big]}$
Coefficient of $\frac{1}{\text{x}}$ in the given expansion = Coefficient of 1 in $(1+\text{x})^{\text{n}}$ × Coefficient of $\frac{1}{\text{x}}$
$={^\text{n}}\text{C}_{\text{0}}\times{^\text{n}}\text{C}_{\text{1}}+{^\text{n}}\text{C}_{\text{1}}\times{^\text{n}}\text{C}_{\text{2}}$
$=\text{n}+\text{n}\times\frac{\text{n}!}{2(\text{n}-2)!}$
$=\text{n}+\text{n}\frac{\text{n}(\text{n}-1)}{2}$
View full question & answer→MCQ 1141 Mark
How many terms are there in the expansion of (4x + 7y)10 + (4x - 7y)10?
MCQ 1151 Mark
The number of terms that are integers in the binomial expansion of $\big(\sqrt{7}+\sqrt[3]{5}\big)^{35}$ is:
AnswerThe general term in the given expansion $\big(\sqrt{7}+\sqrt[3]{5}\big)^{35}$ is $^{35}\text{C}_\text{r}7\frac{35-\text{r}}{2}.5\frac{\text{r}}{3},$
If r is a multiple of 3 and 35 - r is a multiple of 2 then the terms are integers,
$\therefore$ r = 3, 9, 15, 21, 27, 33 which are six values.
View full question & answer→MCQ 1161 Mark
The value of $\sum\limits^\text{n}_{\text{r}=0}\text{a}_{2\text{r}-1}$ is:
- A
$9^n-1$
- ✓
$9^n+1$
- C
$9^n-2$
- D
$9^n+2$
AnswerCorrect option: B. $9^n+1$
- $9^n+1$
Solution:
$\left(1+4 x+4 x^2\right)^n=a_0+a_1 x+a_2 x^2+\ldots\left(a_{2 n} x^{2 n}\right)$
Substituting $x=1$ we get
$9^n=a_0+a_1+a_2+\ldots\left(a_{2 n}\right)$
Substituting $x=-1$ we get
$1=a_0-a_1+a_2-a 3 \ldots\left(a_{2 n}\right)$
Adding both we get
$2\left(a_0+a_2+a_4+\ldots a^{2 n}\right)=9^n+1$
Hence
$\sum^{\text{n}}_{\text{k}=0}\text{a}_{2\text{k}}=9^{\text{n}}+1$ View full question & answer→MCQ 1171 Mark
Find the middle term in the expansion of $\Big(\frac{2\text{x}}{3}+\frac{3}{2\text{x}}\Big)^{10}.$
Answer
- 252
Solution
The middle term will be the 6th term.
It will also be the only term independent of x.
Hence the coefficient will be
$T_{5+1} = ^{10}C_{5}$
$=\frac{10!}{5!(5!)}$
= 252 View full question & answer→MCQ 1181 Mark
The sum of the coefficient in the expansion of $(x + y)^n$ is 4096. The greatest coefficient in the expansion is:
Answer
- 924
Solution:
$(x + y)^n,$ Sum of coefficient = $4096$
When x = y = 1, if n = 12
$\Rightarrow (1 + 1)^{12} = 2^{12} = 4096$
$\Rightarrow$ Hence, greatest coefficient
${^{\text{n}}}\text{C}_{\frac{\text{n}}{2}}={^{12}}\text{C}_{6}=\frac{12!}{6!6!}=924$
Hence, this is the answer. View full question & answer→MCQ 1191 Mark
If the coefficients of 2nd, 3rd and the 4th terms in the expansion of (1 + x)n are in A.P., then value of n is:
MCQ 1201 Mark
The sum of the coefficients in the expansion of $(1-x)^{10}$
Answer
- 0
Solution:
$(1-x)^{10}=1-{ }^{10} C_{1 x}+{ }^{10} C_2 x^2+\ldots{ }^{10} \mathrm{C} 1_0 x^{10}$
Substituting $x=1$, we get sum of coefficients as
$1-{ }^{10} C_1+{ }^{10} C_2+\ldots{ }^{10} C_{10}$
$=(1-1)^{10}$
$=0$ View full question & answer→MCQ 1211 Mark
The coefficient of $x^{−17}$ in the expansion of $\Big(\text{x}^{4}-\frac{1}{\text{x}^{3}}\Big)^{15}$ is:
Answer
- -1365
Solution:
Suppose the $(r + 1)^{th}$ term in the given expansion contains the coefficient of $x^{−17}$.
Then, we have
$\text{T}_{\text{r}+1}={^\text{15}}\text{C}_{\text{r}}(\text{x}^{4})^{15-\text{r}}\Big(\frac{-1}{\text{x}^{3}}\Big)^{\text{r}}$
$\Rightarrow (1)^{\text{r}}\ {^\text{15}}\text{C}_{\text{r}-1}\ \text{x}^{60-4\text{r}-3\text{r}}$
For this term to contain $x^{-17}$, we mst have
$60-7\text{r}=-17$
$\Rightarrow 7\text{r}=77$
$\Rightarrow \text{r}=11$
$\therefore$ Required coefficient $=(-1)^{11}\ {^\text{15}}\text{C}_{\text{11}}=-\frac{15\times14\times13\times12}{4\times3\times2}=-1365$ View full question & answer→MCQ 1221 Mark
The coefficient of $x-12$ in the expansion of $\left(\frac{x+y}{x^3}\right)^{20}$ is:
- A
${ }^{20} \mathrm{C}_8$
- ✓
${ }^{20} \mathrm{C}_8 \mathrm{y}^8$
- C
${ }^{20} \mathrm{C}_{12}$
- D
${ }^{20} \mathrm{C}_{12} \mathrm{y} 12$
AnswerCorrect option: B. ${ }^{20} \mathrm{C}_8 \mathrm{y}^8$
- ${ }^{20} \mathrm{C}_8 \mathrm{y}^8$
View full question & answer→MCQ 1231 Mark
$(\sqrt{3}+1)^{5}-(\sqrt{3}+1)^{5}=$
AnswerIn the above binomial expansion, the terms at the odd position will get eliminated.
We would be left with
$2({^{5}}\text{C}_{1}(\sqrt{3})^{4}+{^5}\text{C}_{3}(\sqrt{3})^{2}+{^5}\text{C}_{5})$
$=2(5(3^2)+10(3)+1)$
$=2(45+30+1)$
$=2(76)$
$=152$
View full question & answer→MCQ 1241 Mark
The term independent of x in the expansion of $\Big(9\text{x}-\frac{1}{3\sqrt{{x}}2}\Big)18,$ x > 0 , is ‘a’ times the corresponding binomial coefficient. Then ‘a’ is:
- A
$3$
- B
$\frac{1}{3}$
- C
$-\frac{1}{3}$
- ✓
$\text{None of these}$
AnswerCorrect option: D. $\text{None of these}$
View full question & answer→MCQ 1251 Mark
Find the sum of the series $3.^{n}C_0 - 8.^nC_1 + 13.^{n}C_2 - 18.^nC_3 + … + (n + 1)$ terms.
Answer
- 0
Solution:
Let n = 2
Hence the above expression is reduced to
3(1) - 8(2) + 13(1)
= 16 - 16 = 0
Let n = 3
3(1) - 8(3) + 13(3) - 18(1)
= 42 - 42 = 0
Hence the sum of the series for n > 1 is 0.
View full question & answer→MCQ 1261 Mark
In the expansion of $\Big(\sqrt[3]4+\frac{1}{\sqrt[4]{6}}\Big)^{20},$
- A
The number of rational terms = 4
- ✓
The number of irrational terms = 19
- C
The middle term is irrational
- D
The number of irrational terms = 17
AnswerCorrect option: B. The number of irrational terms = 19
$\text{T}_{\text{r}+1}={^\text{n}}\text{C}_{\text{r}}4\frac{20-\text{r}}{3}6\frac{-\text{r}}{4}$
$={^\text{n}}\text{C}_{\text{r}}2\frac{40-2\text{r}}{3}-\frac{\text{r}}{4}3-\frac{\text{r}}{4}$
$={^\text{n}}\text{C}_{\text{r}}2\frac{100-11\text{r}}{12}3-\frac{\text{r}}{4}$
There are total of 21 terms.
Hence, we get rational terms for r = 20, 8
Hence there are in total 21 - 2 = 19 irrational terms.
The middle term is at r = 10 which is irrational.
View full question & answer→MCQ 1271 Mark
For $2\leq\text{r}\leq\text{n},$ $\Big(\frac{\text{n}+1}{\text{r}}\Big)+\Big(\frac{\text{n}}{\text{r}-1}\Big)+\Big(\frac{\text{n}}{\text{r}-2}\Big)$is equal to-
- A
$\Big(\frac{\text{n}+1}{\text{r}}\Big)$
- B
$2\Big(\frac{\text{n}+1}{\text{r}-1}\Big)$
- C
$2\Big(\frac{\text{n}+2}{\text{r}}\Big)$
- ✓
$\Big(\frac{\text{n}+2}{\text{r}}\Big)$
AnswerCorrect option: D. $\Big(\frac{\text{n}+2}{\text{r}}\Big)$
- $\Big(\frac{\text{n}+2}{\text{r}}\Big)$
Solution:
${ }^{n+1} C_r+{ }^n C_{r-1}+{ }^n C_{r-2}$
$={ }^{n+1} C_r+{ }^{n+1} C_{r-1}$
$={ }^{n+2} C_r$ View full question & answer→MCQ 1281 Mark
The number of terms in the expansion of (1 + x)21 is:
AnswerThe number of terms in the expansion is one more than n i.e., n + 1
So, here n = 21
The number of terms in the expansion (1 + x)21 = 21 + 1 = 22.
View full question & answer→MCQ 1291 Mark
Choose the correct answer. The total number of terms in the expansion of $(x+a)^{100}+(x-a)^{100}$ after simplification is:
Answer
- 51.
Solution:
Number of terms in the expansion of $(x+a)^{100}=101$
Number of terms in the expansion of $(x-a)^{100}=101$
Now 50 terms of expansion will cancel out with negative 50 terms of $(x-a)^{100}$
So, the remaining 51 terms of first expansion will be added to 51 terms of other.
Therefore, the number of terms $=51$
Hence, the correct option is (c). View full question & answer→MCQ 1301 Mark
If the middle term of $\big(\frac{1}{\text{x}}+\text{x}\sin\text{x}\big)10$ is equal to $7.\frac{7}{8},$ then value of x is:
- A
$2\text{n}\sqcap+\frac{\sqcap}{6}$
- B
$\text{n}\sqcap+\frac{\sqcap}{6}$
- ✓
$\text{n}\sqcap+(-1)^\text{n}\frac{\sqcap}{6}$
- D
$\text{n}\sqcap$
AnswerCorrect option: C. $\text{n}\sqcap+(-1)^\text{n}\frac{\sqcap}{6}$
- $\text{n}\sqcap+(-1)^\text{n}\frac{\sqcap}{6}$
View full question & answer→MCQ 1311 Mark
The coefficient of the middle term in the expansion of (2 + 3x)4 is:
Answer
- 216
Solution:
If the exponent of the expression is n, then the total number of terms is n + 1.
Hence, the total number of terms is 4 + 1 = 5.
Hence, the middle term is the 3rd term.
Therefore, $\mathrm{T}_3={ }^4 \mathrm{C}_2 \times(2)^2 \times(3 \mathrm{x})^2$
$T_3=(6) \times(4) \times\left(9 x^2\right)$
$T_3=216 x^2 .$
Therefore, the coefficient of the middle term is 216. View full question & answer→MCQ 1321 Mark
In the expansion of $\Big(7^{\frac{1}{3}}+11^{\frac{1}{9}}\Big)^{5832},$ the number of terms free from radicals is:
AnswerTotal number of integral terms are
$\frac{5832}{\text{L}.\text{C}.\text{M}(3,9)}+1$
$=\frac{5832}{9}+1$
$=648+1$
$=649$
View full question & answer→MCQ 1331 Mark
In the expansion of $\Big(\frac{1}{2}\text{x}^{\frac{1}{3}}+\text{x}^{\frac{-1}{5}}\Big)^{8},$ the term independent of x is:
- A
$\text{T}_{5}$
- ✓
$\text{T}_{6}$
- C
$\text{T}_{7}$
- D
$\text{T}_{8}$
AnswerCorrect option: B. $\text{T}_{6}$
- $\text{T}_{6}$
Solution:
Suppose the $(r + 1)^{th}$ term in the given expansion is independent of x.
Thus, we have
$\text{T}_{\text{r}+1}={^\text{8}}\text{C}_{\text{r}}\Big(\frac{1}{2}\text{x}^{\frac{1}{3}}\Big)^{8-\text{r}}\Big(\text{x}^{\frac{-1}{5}}\Big)^{\text{r}}$
$={^\text{8}}\text{C}_{\text{r}}\frac{1}{2^{8-\text{r}}}\ \text{x}^{\frac{8-\text{r}}{3}-\frac{\text{r}}{5}}$
For this term to be independent of x, we must have
$\frac{8-\text{r}}{3}-\frac{\text{r}}{5}=0$
$\Rightarrow 40-5\text{r}-3\text{r}=0$
$\Rightarrow \text{r}=5$
Hence, the required term is the 6th term, i.e. $\text{T}_{6}$ View full question & answer→MCQ 1341 Mark
The largest term in the expansion of (3 + 2x)50, when $\text{x}=\frac{1}{5}$ is:
Answer
- 6th term
Solution:
The greatest term in the expansion of (x + y)n is the kth term. Where $\text{k}=\frac{[(\text{n}+1)\text{y}]}{[\text{x}+\text{y}] }...(1)$
On comparing the given expression with the general form, x = 3, y = 2x, n = 50
Now, substitute the values in the given expression, we get
Hence, $k^{th}$ term $= \frac{\big[(50+1)(2\text{x})\big]}{[3+2\text{x}]}$
When $\text{x}=\frac{1}{5}$
$K^{th}$ term $ =\frac{\Big[(51)\big(2\big(\frac{1}{5}\big)\big)\Big]}{\Big[3+2\big(\frac{1}{5}\big)\Big]}=6$
Hence, the 6th term is the largest term in the expansion of (3 + 2x)50, when $\text{x}=\frac{1}{5}.$ View full question & answer→MCQ 1351 Mark
The coefficient of the term independent of x in the expansion of $\Big(\frac{\sqrt{\text{x}}}{3}+\frac{3}{2\text{x}^2}\Big)10$ is:
- ✓
$\frac{5}{4}$
- B
$\frac{7}{4}$
- C
$\frac{9}{4}$
- D
$\text{None of these}$
AnswerCorrect option: A. $\frac{5}{4}$
View full question & answer→MCQ 1361 Mark
If sum of all the coefficients in the expansion of $\Big(\text{x}^{\frac{3}{2}}+\text{x}^{\frac{1}{3}}\Big)^{\text{n}}$ is 128, then the coefficient of $x^5$ is:
Answer
- 35
Solution:
Substituting $\mathrm{x}=1$, we get the sum of the coefficients as
$(2)^n=128$
$\therefore n=7$
Hence writing the general term, we get
$\mathrm{T}_{\mathrm{r}+1}={ }^7 \mathrm{C}_{\mathrm{r}} \mathrm{x} \frac{63-11 \mathrm{r}}{6}$
Hence for the coefficient of $x^5$
$63-11 r=6(5)$
$63-11 r=30$
$33-11 r=0$
$\therefore r=3$
Hence coefficient is ${ }^7 C_3=35$. View full question & answer→MCQ 1371 Mark
Choose the correct answer.The coefficient of $x^n$ in the expansion of $(1 + x)^{2n}$ and $(1 + x)^{2n - 1}$ are in the ratio.
Hint: $^{2\text{n}}\text{C}_\text{n} : \ ^{2\text{n} - 1}\text{C}_\text{n}$
Answer
- 2 : 1.
Solution:
General Term $\text{T}_{\text{r}+1}=\ ^\text{n}\text{C}_\text{r}\text{x}^{\text{n}-\text{r}}\text{y}^\text{r}$
In the expansion of $(1 + x)^{2n}$, we get $\text{T}_{\text{r}+1}=\ ^{2\text{n}}\text{C}_\text{r}\text{x}^\text{r}$
To get the coefficient of $x^n$, put r = n
$\therefore$ Coefficient of $\text{x}^\text{n}=\ ^{\text{2n}}\text{C}_\text{n}$
In the expansion of $(1+\text{x})^{2\text{n}-1},$ we get $\text{T}_{\text{r}+1}=\ ^{2\text{n}-1}\text{C}_\text{r}\text{x}^\text{r}$
$\therefore$ Coefficient of $\text{x}^\text{n}\ \ \text{is}=\ ^{\text{2n}-1}\text{C}_{\text{n}-1}$
The required ratio is $\frac{^{\text{2n}}\text{C}_{\text{n}-1}}{^{\text{2n}-1}\text{C}_{\text{n}-1}}$
$=\frac{\frac{2\text{n}!}{\text{n}!(\text{n}!)}}{\frac{(2\text{n}-1)!}{(\text{n}-1)!(2\text{n}-1-\text{n}+1)!}}=\frac{\frac{2\text{n}!}{\text{n}!.\text{n}!}}{\frac{(2\text{n}-1)!}{(\text{n}-1)!(\text{n}!)}}$
$=\frac{2\text{n}!}{\text{n}!\text{n}!}\times\frac{(\text{n}-\text{n})!\cdot\text{n}!}{(2\text{n}-1)!}=\frac{2\text{n}(2\text{n}-1)!}{\text{n}!\text{n}(\text{n}-1)!}\times\frac{(\text{n}-1)!\cdot\text{n}!}{(2\text{n}-1)!}$
$=\frac{2}{1}=2:1$
Hence, the correct option is (d). View full question & answer→MCQ 1381 Mark
$r$ and $n$ are positive integers $r>1, n>2$ and coefficient of $(r+2)$ th term and $3 r$ th term in the expansion of $(1+x)^{2 n}$ are equal, then $n$ equals:
- A
$3 r$
- B
$3 r+1$
- ✓
$2 r$
- D
$2 \mathrm{r}+1$
View full question & answer→MCQ 1391 Mark
Find the coefficient of $\frac{1}{\text{y}^{2}}$ in $\Big(\text{y}+\frac{\text{c}}{\text{y}^{2}}\Big)^{10}.$
- ✓
$210 c^4$
- B
$210 c^5$
- C
$120 c^3$
- D
$120 c^4$
AnswerCorrect option: A. $210 c^4$
- $210 c^4$
Solution:
$\Big(\text{y}+\frac{\text{c}}{\text{y}^{2}}\Big)^{10}$
$\text{T}_{\text{r}+1}={^{10}}\text{C}_{\text{r}}\text{y}^{10-3\text{r}}\text{c}^{\text{r}}$
Hence for $y^{-2}$
$10-3 r=-2$
$12=3 r$
$r=4$
Coefficient will be
${ }^{10} \mathrm{C}_4 \mathrm{c}^4$
$=210 \mathrm{c}^4$ View full question & answer→MCQ 1401 Mark
If the coefficients of $x^2$ and $x^3$ in the expansion of $(3+\text{ax})^{9}$ are the same, then the value of a is:
- A
$-\frac{7}{9}$
- B
$-\frac{9}{7}$
- C
$\frac{7}{9}$
- ✓
$\frac{9}{7}$
AnswerCorrect option: D. $\frac{9}{7}$
- $\frac{9}{7}$
Solution:
Coefficients of $x^2$ = Coefficients of $x^3$
${^\text{9}}\text{C}_{\text{2}}\times3^{9-2}\text{a}^{2}={^\text{9}}\text{C}_{\text{3}}\times3^{9-3}\ \text{a}^{3}$
$\Rightarrow \text{a}=\frac{{^\text{9}}\text{C}_{\text{2}}}{{^\text{9}}\text{C}_{\text{3}}}\times3$
$=\frac{9!\times3!\times6!\times3}{2!\times7!\times9!}$
$=\frac{9}{7}$ View full question & answer→MCQ 1411 Mark
The coefficient of the middle term in the expansion of (2 + 3x)4 is:
MCQ 1421 Mark
The middle term in the expansion of $\Big(\frac{\text{a}}{\text{x}}+\text{bx}\Big)^{12}$ is:
- ✓
$924 a^6 b^6$
- B
$924 a^6 b^5$
- C
$924 a^5 b^5$
- D
$924 a^5 b^6$
AnswerCorrect option: A. $924 a^6 b^6$
- $924 a^6 b^6$
Solution:
The middle term will be the 7th term.
Hence $\text{T}_{6+1}=^{12}\text{C}_6\big(\frac{\text{a}}{\text{x}}\big)^6(\text{bx})^6=924\text{a}^6\text{b}^6$ View full question & answer→MCQ 1431 Mark
In the expansion of $(1+x)^n \cdot(1+y)^n \cdot(1+z)^n$ the sum of the coefficients of the terms of degree $r$ is:
- A
$\left({ }^n \mathrm{C}_r\right)^3$
- ✓
${ }^{3 n} \mathrm{C}_{\mathrm{r}}$
- C
${ }^{3 \times n} C_r$
- D
${ }^n \mathrm{C}_{3 \mathrm{r}}$
AnswerCorrect option: B. ${ }^{3 n} \mathrm{C}_{\mathrm{r}}$
- ${ }^{3 n} \mathrm{C}_{\mathrm{r}}$
Solution:
The given expression contains 3n factors
Using combination to choose r brackets out of 3n brackets for a term of degree r, we get
${ }^{3 n} \mathrm{C}_{\mathrm{r}}$ View full question & answer→MCQ 1441 Mark
The middle term in the expansion of $\Big(\text{x}+\frac{1}{\text{x}}\Big)^{10},$ is:
AnswerCorrect option: B. ${ }^{10} \mathrm{C}_5$
- ${ }^{10} \mathrm{C}_5$
Solution
The middle term would be the 6th term.
Hence
$T_{5+1} = ^{10}C_5$. View full question & answer→MCQ 1451 Mark
$(x-1)^4+4(x-1)^3+6(x-1)^2+4(x-1)+1=$
Answer
- $x^4$
Solution:
Consider the following identity
$(a+1)^4$
$=\left((a+1)^2\right)^2$
$=\left(a^2+2 a+1\right)^2$
$=a^4+4 a^2+1+4 a^3+4 a+2 a^2$
$=a^4+4 a^3+6 a^2+4 a+1 \ldots \text { (i) }$
Comparing i with the given question we get
$a=(x-1)$
Therefore
$(x-1)^4+4(x-1)^3+6(x-1)^2+4(x-1)+1$
$=(x-1+1)^4 \text { from (i) }$
$=x 4$. View full question & answer→MCQ 1461 Mark
If the coefficent of $x^2$ in the expansion of $(1+x)^m$ is 6 then $m=$ $\qquad$
Answer
- 4
Solution:
The coefficient of $x^r$ in $(1+x)^n$ is ${ }^n C_r$
$\Rightarrow$ the coefficient of $x^2$ in $(1+x)^m$ is ${ }^m C_2=6$
$\Rightarrow\frac{\text{m}(\text{m}-1)}{2}=6$
$\Rightarrow\text{m}^{2}-\text{m}-12=0$
$\Rightarrow\text{m}=4,-3$
$\therefore$ m is a positive number,
so m = 4. View full question & answer→MCQ 1471 Mark
After simplification, what is the number of terms in the expansion of [(3x + y)5]4 - [(3x - y)4]5?
MCQ 1481 Mark
The number of integral terms in$ (\sqrt{3}+\sqrt[8]{2})^{64}$ is-
Answer
- 9
Solution:
The general term of expansion $(x + y)^n$ is $^nC_rx^{n−r}y^r$
So the general term of $ (\sqrt{3}+\sqrt[8]{2})^{64}$ is $ {^{64}}{\text{C}}_{\text{r}}3\frac{64-\text{r}}{2}2\frac{\text{r}}{8}$
For the term to be integer, r must be divided by 8 and 64 - r must be divided by 2
The possible values of r are 0, 8, 16, 24, 32, 40, 48, 56, 64 the number of integral values is 9. View full question & answer→MCQ 1491 Mark
The number of irrational terms in the expansion of $\Big(2^{\frac{1}{5}}+3^{\frac{1}{10}}\Big)^{55}$ is:
AnswerFor the above question $\text{T}_{\text{r}+1}={^{55}}\text{C}_{\text{r2}}11-\frac{\text{r}}{5}3\frac{\text{r}}{10}$
Hence we will have rational terms at r = 0, 10, 20, 30, 40, 50 respectively.
Hence there will be 6 rational terms.
The total number of terms will be
55 + 1
= 56 terms.
Hence the number of irrational terms will be
56 - 6
= 50 terms.
View full question & answer→MCQ 1501 Mark
The coefficient of xp and xq (p and q are positive integers) in the expansion of (1 + x) p + q are:
- ✓
- B
Equal with opposite signs
- C
- D
View full question & answer→MCQ 1511 Mark
The number of irrational terms in the expansion of $\Big(4^{\frac{1}{5}}+7^{\frac{1}{10}}\Big)^{45}$ is:
Answer
- 41
Solution:
The general term $T_{r+1}$ in the given expansion is given by ${^\text{45}}\text{C}_{\text{r}}\Big(4^{\frac{1}{5}}\Big)^{45-\text{r}}\Big(7^{\frac{1}{10}}\Big)^{\text{r}}$
For $T_{r+1}$ to be an integer, we must have $\frac{\text{r}}{5}$ and $\frac{\text{r}}{10}$ as integers i.e. $0\leq\text{r}\leq45$
$\therefore \text{r}=0,10,20,30,40$
Hence, there are 5 rational and 41, i.e. 46 - 5, irrational terms. View full question & answer→MCQ 1521 Mark
Sum of the coefficients in the expansion of $(5 x-4 y)^n$ where $n$ is a positive integer is:
- ✓
$1$
- B
$9^n$
- C
$(-1)^n$
- D
$5^n$
Answer
- $1$
Solution
Substituting $x=1$ and $y=1$ in the above expression we get the sum of the binomial coefficients as
$(5 x-4 y)^n$
$=(5-4)^n \text { (substituting } x=1 \text { and } y=1 \text { ) }$
$=(1)^n$
$=1$ View full question & answer→MCQ 1531 Mark
Choose the correct answer.If A and B are coefficient of $x^n$ in the expansions of $(1 + x)^{2n}$ and $(1 + x)^{2n – 1}$ respectively, then $\frac{\text{A}}{\text{B}}$ equals:
Hint: $\frac{\text{A}}{\text{B}}=\frac{^{2\text{n}}\text{C}_\text{n}}{^{2\text{n}-1}\text{C}_\text{n}}=2$
- A
$1.$
- ✓
$2.$
- C
$\frac{1}{2}.$
- D
$\frac{1}{\text{n}}.$
Answer
- $2.$
Solution:
Given expression is $(1 + x)^{2n}$
$\text{T}_{\text{r}+1}=\ ^{2\text{n}}\text{C}_\text{r}\text{x}^\text{r}$
$\therefore$ Coefficient of $xn = ^{2n}C_n = A$ (Given)
In the expression of $(1 + x)^{2n - 1}$
$\text{T}_{\text{r}+1}=2^{\text{n}-1}\text{C}_\text{x}=\text{B}$ Given
So, $\frac{\text{A}}{\text{B}}=\frac{^{2\text{n}}\text{C}_\text{n}}{^{2\text{n}-1}\text{C}_\text{n}}=\frac{^{2\text{n}}\text{C}_\text{n}}{^{2\text{n}-1}\text{C}_\text{n}}=\frac{2}{1}$
Hence, the correct option is (b). View full question & answer→MCQ 1541 Mark
The coefficient of $x^5$ in the expansion of $\left(1+x^2\right)^5(1+x)^4$ is:
Answer
- 60
Solution:
Given equationnis $(1+x)^4\left(1+x^2\right)^5$
$=\left(1+4 x+6 x^2+4 x^3+x^4\right)\left(1+5 x^2+10 x^4+10 x^6+5 x^8+x^{10}\right) \ldots(i)$
Hence the coefficient of $x^5$ from Eq (i) will be.
$4(10)+4(5)$
$=40+20$
$=60$ View full question & answer→MCQ 1551 Mark
If the fourth term of the binomial expansion of $\Big(\text{px}+\big(\frac{1}{\text{x}}\big)\Big)^\text{n}$ is $\frac{5}{2},$ then:
- A
$\text{n}=6, \text{p}=6$
- B
$\text{n}=8, \text{p}=6$
- C
$\text{n}=8, \text{p}=\frac{1}{2}$
- ✓
$\text{n}=6, \text{p}=\frac{1}{2}$
AnswerCorrect option: D. $\text{n}=6, \text{p}=\frac{1}{2}$
Given: $\Big(\text{px}+\big(\frac{1}{\text{x}}\big)\Big)^\text{n}$
Hence, the fourth term, $ \text{T}_{\text{r}+1}={^{\text{n}}}{\text{C}}_{3}(\text{px})^{\text{n}-3}\big(\frac{1}{\text{x}}\big)3$
Given that the fourth term of the binomial expansion of $\Big(\text{px}+\big(\frac{1}{\text{x}}\big)\Big)^\text{n}$ is $\frac{5}{2},$ which is independent of x.
Hence, $\big(\frac{5}{2}\big)={^\text{n}}{\text{C}}_{3}(\text{px})^{\text{n}-3}\big(\frac{1}{\text{x}}\big)^3 …(1)$
On solving this, we get n = 6.
Now, substitute n = 6 in $\big(\frac{5}{2}\big)=6{ \ ^\text{n}}{\text{C}}_{3}(\text{p})3$
$20\text{p}^3=\frac{5}{2}$
$\text{p}^3=\frac{5}{2}$
$\text{p}=\frac{1}{2}$
Therefore, n = 6 and $\text{p}=\frac{1}{2}.$
View full question & answer→MCQ 1561 Mark
If tr is the rth term in the expansion of (1+ x)101, then what is the ratio $\frac{\text{t}20}{\text{t}19}$ equal to?
- A
$\frac{20\text{x}}{19}$
- B
$83\text{x}$
- C
$19\text{x}$
- ✓
$\frac{83\text{x}}{19}$
AnswerCorrect option: D. $\frac{83\text{x}}{19}$
View full question & answer→MCQ 1571 Mark
Number of irrational terms in the expansion of $\Big(5^{\frac{1}{6}}+2^{\frac{1}{8}}\Big)^{100}$ is:
Answer$\text{T}_{\text{r}+1}=^{100}\text{C}_\text{r}5^{\frac{(\text{r}-100)}{6}}2^{\frac{\text{r}}{8}}$
Hence we get rational terms when
r = 8k where k is an integer and $\frac{8\text{k}-100}{6}$ is an integer
r = 16, 40, 64, 88
Hence we get in total 4 rational terms.
However, total number of terms will be 101
Hence total number of irrational terms is 101 - 4
= 97 terms.
View full question & answer→MCQ 1581 Mark
What is the middle term in the expansion of $\Big(\frac{\text{x}\sqrt{\text{y}}}{3}-\frac{3}{\text{y}\sqrt{\text{x}}}\Big)^{12}$?
- A
$C(12,7) x^3 y^{-3}$
- B
$C(12,6) x^{-3} y^3$
- C
$C(12,7) x^{-3} y^3$
- ✓
$C(12,6) x^3 y^{-3}$
AnswerCorrect option: D. $C(12,6) x^3 y^{-3}$
View full question & answer→MCQ 1591 Mark
If the coefficients of 2nd, 3rd and the 4th terms in the expansion of $(1 + x)^n$ are in A.P, then value of n is:
Answer
- 7
Solution:
For the terms to be in A.P, it must follow
$(n-2 r)^2=n+2$
In the above case $r=2$
Substituting in the equation, we get
$(n-4)^2=n+2$
$n^2-8 n+16=n+2$
$n^2-9 n+14=0$
$(n-2)(n-7)=0$
$n=2 \text { and } n=7$
However for $n=2$ we will have only 3 terms.
Hence the required answer is 7 . View full question & answer→MCQ 1601 Mark
If rth term in the expansion of $\Big(2\text{x}^{2}-\frac{1}{\text{x}}\Big)^{12}$ is without x, then r is equal to:
Answerrth term in the given expansion is ${^\text{20}}\text{C}_{\text{r}-1}\Big(2\text{x}^{2}\Big)^{12-\text{r}+1}\Big(\frac{-1}{\text{x}}\Big)^{\text{r}-1}$
$=(-1)^{\text{r}-1}\ {^\text{20}}\text{C}_{\text{r}-1}\ 2^{13-\text{r}}\ \text{x}^{26-2\text{r}-\text{r}+1}$
For this term to be independent of x, we must have,
$27-3\text{r}=0$
$\Rightarrow \text{r}=9$
Hence, the term in the expansion is independent.
View full question & answer→MCQ 1611 Mark
The middle term in the expansion of$\Big(1+\frac{1}{\text{x}^{2}}\Big)\big(1+\text{x}^{2}\big)^{\text{n}}$ is:
AnswerCorrect option: C. ${ }^{2 n} \mathrm{Cn}$
View full question & answer→MCQ 1621 Mark
Choose the correct answer. If the middle term of $\Big(\frac{1}{\text{x}}+\text{x}\sin\text{x}\Big)^{10}$ is equal to $7\frac{7}{8},$ then value of x is:
Hint: $\text{T}_6=\ ^{10}\text{C}_5\frac{1}{\text{x}^5}.\text{x}^5\ \sin^5\text{x}=\frac{63}{8}\Rightarrow\sin^5\text{x}=\frac{1}{2^5}\sin\frac{1}{2}$ $\Rightarrow\text{x}=\text{n}\pi+(-1)^\text{n}\frac{\pi}{6}$
- A
$2\text{n}\pi+\frac{\pi}{6}.$
- B
$\text{n}\pi+\frac{\pi}{6}.$
- ✓
$\text{n}\pi+(-1)^\text{n}\frac{\pi}{6}.$
- D
$\text{n}\pi+(-1)^\text{n}\frac{\pi}{3}.$
AnswerCorrect option: C. $\text{n}\pi+(-1)^\text{n}\frac{\pi}{6}.$
- $\text{n}\pi+(-1)^\text{n}\frac{\pi}{6}.$
Solution:
Given expression is $\Big(\frac{1}{\text{x}}+\text{x}\sin\text{x}\Big)^{10}$
Since, n = 10 (even), so there is only one middle term which is, $6^{th}$ term.
$\therefore\text{T}_6=\text{T}_{5+1}=\ ^{10}\text{C}_5\Big(\frac{1}{\text{x}}\Big)^{10-5}(\text{x}\sin\text{x})^5$
$\Rightarrow\frac{63}{8}=\ ^{10}\text{C}_5\sin^5\text{x}$ (given)
$\Rightarrow\frac{63}{8 }=252\times\sin^5\text{x}\Rightarrow\sin^5\text{x}=\frac{1}{32}$ $\Rightarrow\sin\text{x}=\frac{1}{2}\Rightarrow\sin\text{x}=\sin\frac{\pi}{6}$
$\Rightarrow\text{x}=\text{n}\pi+(-1)^\text{n}\frac{\pi}{6}$ View full question & answer→MCQ 1631 Mark
The number of rational terms in the expansion of $(\sqrt3+\sqrt[4]{5})^{124}$ is:
Answer
- 32
Solution:
$\text{T}_\text{r}=^{124}\text{C}_{\text{r}-1}(\sqrt3)^{125-\text{r}}(\sqrt[4]{5})^{\text{r}-1}$
When both the terms are rational, $T_r$ will be rational.
Hence, $\frac{125-\text{r}}{2}$ and $\frac{\text{r}-1}{4}$ both must be integers.
Therefore, r must be of the form 4k + 1, where k is an integer.
There are 125 terms in the expansion.
Hence, k can assume values from 0 to 31.
Hence, there are 32 values of k and 32 rational terms in the expansion. View full question & answer→MCQ 1641 Mark
If in the expansion of $(1+\text{y})^{\text{n}},$ the coefficients of $5^{th}, 6^{th}$ and $7^{th}$ terms are in A.P., then n is equal to:
Answer
- 7, 14
Solution:
Coefficients of the $5^{th}, 6^{th}$ and $7^{th}$ terms in the given expansion are ${^\text{n}}\text{C}_{\text{4}},{^\text{n}}\text{C}_{\text{5}}$ and ${^\text{n}}\text{C}_{\text{6}}.$
These coefficients are in AP.
Thus, we have
$2\ {^\text{n}}\text{C}_{\text{5}}={^\text{n}}\text{C}_{\text{4}}+{^\text{n}}\text{C}_{\text{6}}$
On dividing both sides by ${^\text{n}}\text{C}_{\text{5}},$ we get
$2=\frac{{^\text{n}}\text{C}_{\text{4}}}{{^\text{n}}\text{C}_{\text{5}}}+\frac{{^\text{n}}\text{C}_{\text{6}}}{{^\text{n}}\text{C}_{\text{5}}}$
$\Rightarrow 2=\frac{5}{\text{n}-4}+\frac{\text{n}-5}{6}$
$\Rightarrow 12\text{n}-48=30+\text{n}^{2}-4\text{n}-5\text{n}+20$
$\Rightarrow \text{n}^{2}-21\text{n}+98=0$
$\Rightarrow (\text{n}-14)(\text{n}-7)=0$
$\Rightarrow \text{n}=7,14$ View full question & answer→MCQ 1651 Mark
The fourth term in the expansion of $(x-2 y)^{12}$ is:
- ✓
$-1760 x^9 \times y^3$
- B
$-1670 x^9 \times y^3$
- C
$-7160 x^9 \times y^3$
- D
$-1607 x^9 \times y^3$
AnswerCorrect option: A. $-1760 x^9 \times y^3$
- $-1760 x^9 \times y^3$
Solution:
We know that the general term of an expansion $(a+b) n$ is $T_{r+1}={ }^n C_r a^{n-r}$ br.
Now, we have to find the fourth term in the expansion $(x-2 y)^{12}$
Hence, $r=3, a=x, b=-2 y, n=12$.
Now, substitute the values in the formula, we get
$T_{3+1}={ }^{12} C_3 x^{12-3}(-2 y)^3$.
On solving this, we get
$T_4=-1760 x^9 y^3$. View full question & answer→MCQ 1661 Mark
${ }^5 C_0+2 \cdot{ }^5 C_1+2^{2.5} \cdot{ }^5 C_2+2^3 \cdot{ }^5 C_3+2^4 \cdot 5 \mathrm{C}_4+2^5 \cdot{ }^5 C_5=$
Answer
- 243
Solution
Just calculate it by substituting the values of ${ }^5 \mathrm{C}_1,{ }^5 \mathrm{C}_2$ etc as 5,10 and just add them to get 243. View full question & answer→MCQ 1671 Mark
In the expansion of $\Big(\sqrt{2}+\sqrt[5]{3}\Big)^{120}$ the number of irrational terms is:
AnswerTotal number of rational terms is
$\frac{120}{\text{L}.\text{C}.\text{M}(5,2)}+1$
$\frac{120}{10}+1$
$=13$
Hence total number of irrational terms are
= 121 - 13
= 108
View full question & answer→MCQ 1681 Mark
$(1-\sqrt{2})^{6}=$
- A
$98-70\sqrt{2}$
- ✓
$99-70\sqrt{2}$
- C
$99+70\sqrt{2}$
- D
$98+70\sqrt{2}$
AnswerCorrect option: B. $99-70\sqrt{2}$
$(1-\sqrt{2})^{6}$
$=((1-\sqrt{2})^{2})^{3}$
$=(1+2-2\sqrt{2})^{3}$
$=(3-2\sqrt{2})^{3}$
$=27-16\sqrt{2}-54\sqrt{2}+72$
$=99-70\sqrt{2}.$
View full question & answer→MCQ 1691 Mark
If $\Big(\frac{\text{P}}{\text{q}}\Big)=0$ for p < q p, $\text{q}\in\text{W}$ then $\sum^\limits{\infty}_{\text{r}=0}\big(\frac{\text{n}}{2\text{r}})$
- A
$2^n$
- ✓
$2^{n-1}$
- C
$2^{2 \mathrm{n}-1}$
- D
$2^n C_n$
AnswerCorrect option: B. $2^{n-1}$
- $2^{n-1}$
Solution:
$\sum{^\text{n}}\text{C}_{2\text{r}}$
Is the sum of even odd term in the binomial expansion of $(1 + x)^n$
Hence
$\sum{^\text{n}}\text{C}_{2\text{r}}$will always be equal to $2^{n-1}$. View full question & answer→MCQ 1701 Mark
The coefficient of x - 7 in the expansion of $\Big[\text{ax}2-\frac{1}{\text{bx}}^{2}\Big]11$ will be:
- A
$\frac{462}{\text{b}^5}\times\text{a}^6$
- ✓
$\frac{462\text{a}^2}{\text{b}^6}$
- C
$\frac{462\text{a}^5}{\text{b}^6}$
- D
$\frac{-462\text{a}^6}{\text{b}^5}$
AnswerCorrect option: B. $\frac{462\text{a}^2}{\text{b}^6}$
- $\frac{462\text{a}^2}{\text{b}^6}$
View full question & answer→MCQ 1711 Mark
Choose the correct answer. Given the integers $r>1, n>2$, and coefficients of $(3 r)^{\text {th }}$ and $(r+2)^{n d}$ terms in the binomial expansion of $(1+x)^{2 n}$ are equal, then:
Answer
- n = 2r.
Solution:
The given expression is $(1 + \text{x})^{2\text{n}}$
$\therefore\text{T}_{3\text{r}}=\text{T}_{(3\text{r}-1)+1}=\ ^{2\text{n}}\text{C}_{3\text{r}-1}\ \text{x}^{3\text{r}-1}$
and $\text{T}_{\text{r}+2}=\text{T}_{(\text{r}+1)+1}=\ ^{2\text{n}}\text{C}_{\text{r}+1}\text{x}^{\text{r}+1}$
Given, $^{2\text{n}}\text{C}_{3\text{r}-1}=\ ^{2\text{n}}\text{C}_{\text{r}+1}$
$\Rightarrow3\text{r}-1+\text{r}+1=2\text{n}\ \ [\because\ ^\text{n}\text{C}_\text{x}=\ ^\text{n}\text{C}_\text{y}\Rightarrow\text{x}+\text{y}=\text{n}]$
$\therefore\text{n}=2\text{r}$ View full question & answer→MCQ 1721 Mark
The number of irrational terms in the expansion of $\Big(4^{\frac{1}{5}}+7^{\frac{1}{10}}\Big)^{45}$ is:
AnswerTotal number of terms in the expansion of $\Big(4^{\frac{1}{5}}+7^{\frac{1}{10}}\Big)^{45}$ is 45 + 1 = 46
The general term in the expansion is $\text{T}{\text{r}+1}={^{45}}\text{C}_{\text{r}}\times4\frac{45-\text{r}}{5}\times7^{\frac{\text{r}}{10}}\text{T}_{\text{r}+1}$ is rational if r = 0, 10, 20, 30, 40
$\therefore$ Number of rational terms = 5
$\therefore$ Number of irrational terms = 46 - 5 = 41
View full question & answer→