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BINOMIAL THEOREM · Gujarat Board (GSEB) STD 11 Science (GSEB - English Medium). 17 questions.

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Question 11 Mark
The coefficients of three consecutive terms in the expansion of $(1 + a)^n$ are in the ratio $1 : 7 : 42$. Find $n$.
Answer
Suppose three consecutive terms in the expansion of $(1+a)^n$ are $(r-1)$ th, rth and $(r+1)$ th terms.
$\therefore T_{r-1}=T_{(r-2)+1}={ }^n C_{(r-2)}(1)^{n-r+2}(a)^{r-2}$
$\Rightarrow T_{r-1}={ }^n C_{r-2} a^{r-2} \ldots . \text { (i) }$
$T_r=T_{(r-1)+1}={ }^n C_{r-1}(1)^{n-r+1}(a)^{r-1}$
$\Rightarrow T_r={ }^n C_{r-1} a^{r-1} \ldots \text { (ii) }$
$\text { and } T_{r+1}={ }^n C_r(1)^{n-r} \text { (a) }$
$\Rightarrow T_{r+1}={ }^n C_r a^r \ldots \text {... (iii) }$
Since, coefficients are in the ratio 1 : 7 : 42, so we have
$\frac { \text { Coefficient of } T _ { r - 1 } } { \text { Coefficient of } T _ { r } } = \frac { 1 } { 7 } \Rightarrow \frac { ^ { n } C _ { r - 2 } } { ^ { n } C _ { r - 1 } } = \frac { 1 } { 7 }$
$\Rightarrow$ $\frac { \frac { n ! } { ( r - 2 ) ! ( n - r + 2 ) ! } } { \frac { n ! } { ( r - 1 ) ! ( n - r + 1 ) ! } } = \frac { 1 } { 7 }$
$\Rightarrow$ $\frac { ( r - 1 ) ! ( n - r + 1 ) ! } { ( r - 2 ) ! ( n - r + 2 ) ! } = \frac { 1 } { 7 }$
$\Rightarrow$ $\frac { ( r - 1 ) ( r - 2 ) ! ( n - r + 1 ) ! } { ( r - 2 ) ! ( n - r + 2 ) ( n - r + 1 ) ! } = \frac { 1 } { 7 }$
$\Rightarrow$ $\frac { ( r - 1 ) } { n - r + 2 } = \frac { 1 } { 7 }$
$\Rightarrow$ 7r - 7 = n - r + 2
$\Rightarrow$ 8r - n = 9 ... (iv)
and $\frac { \text { Coefficient of } T _ { r } } { \text { Coefficient of } T _ { r + 1 } } = \frac { 7 } { 42 } \Rightarrow \frac { ^ { n } C _ { r - 1 } } { ^ { n } C _ { r } } = \frac { 7 } { 42 }$
$\Rightarrow$ $\frac { \frac { n ! } { ( r - 1 ) ! ( n - r + 1 ) ! } } { \frac { n ! } { r ! ( n - r ) ! } } = \frac { 1 } { 6 }$
$\Rightarrow$ $\frac { r ! ( n - r ) ! } { ( r - 1 ) ! ( n - r + 1 ) ! } = \frac { 1 } { 6 }$
$\Rightarrow$ $\frac { r ( r - 1 ) ! ( n - r ) ! } { ( r - 1 ) ! ( n - r + 1 ) ( n - r ) ! } = \frac { 1 } { 6 }$
$\Rightarrow$ $\frac { r } { ( n - r + 1 ) } = \frac { 1 } { 6 } \Rightarrow 6 r = n - r + 1$
$\Rightarrow$ 7r - n = 1 ... (v)
On subtracting Eq. (v) from Ew. (iv), we get
r = 8
On putting the value of r in Eq. (iv), we get
8 $\times$ 8 - n = 9 $\Rightarrow$ n = 64 - 9
$\Rightarrow$ n = 55
Hence, the value of n is 55.
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Question 21 Mark
The $2nd, 3rd$ and $4th$ terms in the binomial expansion $(x + a)^n$​​​​​​​ are $240, 720$ and $1080$, respectively. Find the values of x, a and n.
Answer
Given second term, $T_2=240, T_3=720$ and $T_4=1080$
In the expansion of $(x+a)^n$, we get
$T_2=T_{1+1}={ }^n C_1 x^{n-1} a$
$\therefore T_2={ }^n C_1 x^{n-1} a=240 \ldots$
Similarly, $T_3=T_{2+1}={ }^n C_1 x^{n-2} a^2=720 \ldots$
and $T_4=T_{3+1}={ }^n C_1 x^{n-3} a^3=1080 \ldots$
On dividing Eq. (ii) by Eq. (i), we get
$\frac { ^ { n } C _ { 2 } \cdot x ^ { n - 2 } \cdot a ^ { 2 } } { ^ { n } C _ { 1 } \cdot x ^ { n - 1 } \cdot a } = \frac { 720 } { 240 }$
$\Rightarrow \frac { ^ { n } C _ { 2 } } { ^ { n } C _ { 1 } } \cdot x ^ { - 1 } \cdot a = 3 \quad \Rightarrow \frac { \frac { n ! } { 2 ! ( n - 2 ) ! } } { \frac { n ! } { 1 ! ( n - 1 ) ! } } \times \frac { a } { x } = 3$
$\Rightarrow \frac { ( n - 1 ) ! } { 2 ! ( n - 2 ) ! } \cdot \frac { a } { x } = 3 \Rightarrow \frac { ( n - 1 ) ( n - 2 ) ! } { 2 \times 1 ( n - 2 ) ! } \cdot \frac { a } { x } = 3$
$\Rightarrow ( n - 1 ) \cdot \frac { a } { x } = 6 \quad \Rightarrow \quad \frac { a } { x } = \frac { 6 } { n - 1 }$ ... (iv)
On dividing Eq. (iii) by Eq. (ii), we get
$\frac { ^ { n } C _ { 3 } \cdot x ^ { n - 3 } \cdot a ^ { 3 } } { ^ { n } C _ { 2 } \cdot x ^ { n - 2 } \cdot a ^ { 2 } } = \frac { 1080 } { 720 }$
$\Rightarrow$ $ \frac { ^ { n }C _ { 3 } } { ^ { n } C _ { 2 } } \cdot x ^ { - 1 } \cdot a = \frac { 3 } { 2 }$
$\Rightarrow$ $\frac { \frac { n ! } { 3 ! ( n - 3 ) ! } } { \frac { n ! } { 2 ! ( n - 2 ) ! } } \cdot \frac { a } { x } = \frac { 3 } { 2 }$
$\Rightarrow$ $\frac { 2 ! ( n - 2 ) ! } { 3 ! ( n - 3 ) ! } \cdot \left( \frac { a } { x } \right) = \frac { 3 } { 2 }$
$\Rightarrow$ $\frac { 2 \times 1 ( n - 2 ) ( n - 3 ) ! } { 3 \times 2 \times 1 ( n - 3 ) ! } \left( \frac { a } { x } \right) = \frac { 3 } { 2 }$
$\Rightarrow$ $\frac { ( n - 2 ) } { 3 } \left( \frac { a } { x } \right) = \frac { 3 } { 2 }$
$\Rightarrow$ $\frac { a } { x } = \frac { 9 } { 2 ( n - 2 ) }$ ... (v)
On equating Eqs. (iv) and (v), we get
$\frac { 6 } { n - 1 } = \frac { 9 } { 2 ( n - 2 ) }$
$\Rightarrow$ 12 (n - 2) = 9 (n - 1)
$\Rightarrow$ 12n - 24 = 9n - 9 $\Rightarrow$ 3n = 15
$\therefore$ n = 5
On putting n = 5 in Eq. (i), we get
$T_2 = ^5C_1x^{5-1} a = 240 \Rightarrow 5 a x^4 = 240 ... (vi)$
From Eq. (iv), we get
$\frac { a } { x } = \frac { 6 } { n - 1 } = \frac { 6 } { 5 - 1 } = \frac { 6 } { 4 } = \frac { 3 } { 2 }$
$\Rightarrow$ a = $\frac { 3 } { 2 } x$ ... (vii)
On putting the value of a = $\frac { 3 } { 2 } x$ in Eq. (vi), we get
$5 x^4 \times$ $\frac { 3 } { 2 } x$ = 240
$\Rightarrow$ $\frac { 15 } { 2 } \cdot x ^ { 5 }$ = 240
$\Rightarrow$ x^5 = $\frac { 240 \times 2 } { 15 }$
$\Rightarrow x^5 = 32 = 2^5$
$\therefore x = 2$
On putting the value of x = 2 in Eq. (vii), we get
a = $\frac { 3 } { 2 } \cdot x$ = $\frac { 3 } { 2 } \times 2$
$\Rightarrow$ a = 3
Hence, we get the values x = 2, a = 3 and n = 5.
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Question 31 Mark
Find the coefficient of $x^6y^3$ in the expansion of $(x + 2y)^9$.
Answer
In the expansion of $(x+2 y)^9$, the general term is
$T_{r+1}={ }^9 C_r x^{9-r}(2 y)^r\left[\because \text { for }(x+a)^n, T_{r+1}={ }^n C_r x^{n-r} a^r\right]$
$={ }^9 C_r x^{9-r} \times 2^r \times y^r={ }^9 C_r \times 2^r \times x^{9-r} \times y^r \ldots \text { (i) }$
This will contain $x^6 y^3$, if $9-r=6$ or $r=3$.
On putting $r=3$ in Eq. (i), we get
$T_{3+1}={ }^9 \mathrm{C}_3 2^3 \times \mathrm{x}^{9-3} \times \mathrm{y}^3$
$={ }^9 \mathrm{C}_3 \times 8 \times \mathrm{x}^6 \times \mathrm{y}^3=\frac{9!}{316!} \times 8 \times \mathrm{x}^6 \times \mathrm{y}^3$
$\therefore \text { Coefficient of } \mathrm{x}^6 \mathrm{y}^3=\frac{9 \times 8 \times 7}{3 \times 2 \times 1} \times 8=672$
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Question 41 Mark
Show that the middle term in the expansion of $(1 + x)^{2n}$ is $\frac { 1 \cdot 3 \cdot 5 \ldots ( 2 n - 1 ) } { n ! } 2^n x^n$, where n is a positive integer.
Answer
As 2n is even, the middle term in the expansion of $(1 + x)^{2n}$ is $\left( \frac { 2 n } { 2 } + 1 \right)$th i.e., (n + 1)th term, which is given by
$T_{n+1} = ^{2n}C_n(1)^{2n-n}(x)^n = ^{2n}C_nx^n = \frac { ( 2 n ) ! } { n ! ( 2 n - n ) ! } \cdot x ^ { n }$
= $\frac { ( 2 n ) ! } { n ! n ! } \cdot x ^ { n }$
= $\frac { 2 n ( 2 n - 1 ) ( 2 n - 2 ) \cdots 4 \cdot 3 \cdot 2 \cdot 1 } { n ! n ! } \cdot x ^ { n }$
= $\frac { 1 \cdot 2 \cdot 3 \cdot 4 \cdots ( 2 n - 2 ) ( 2 n - 1 ) ( 2 n ) } { n ! n ! } \cdot x ^ { n }$
= $\frac { [ 1 \cdot 3 \cdot 5 \ldots ( 2 n - 1 ) ] [ 2 \cdot 4 \cdot 6 \cdots ( 2 n ) ] } { n ! n ! } \cdot x ^ { n }$
= $\frac { [ 1 \cdot 3 \cdot 5 \cdots ( 2 n - 1 ) ] [ 2 \cdot ( 2 \cdot 2 ) \cdot ( 2 \cdot 3 ) \cdots ( 2 \cdot n ) ] \cdot x ^ { n } } { n ! n ! }$
= $\frac { [ 1 \cdot 3 \cdot 5 \cdots ( 2 n - 1 ) ] 2 ^ { n } [ 1 \cdot 2 \cdot 3 \cdots n ] } { n ! n ! } \cdot x ^ { n }$
= $\frac { [ 1 \cdot 3 \cdot 5 \cdots ( 2 n - 1 ) ] 2 ^ { n } \cdot n ! } { n ! n ! } \cdot x ^ { n }$
= $\frac { [ 1 \cdot 3 \cdot 5 \cdots ( 2 n - 1 ) ] \cdot 2 ^ { n } } { n ! } \cdot x ^ { n }$
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Question 51 Mark
Find a if the $17^{th}$ and $18^{th}$ terms of the expansion $(2 + a)^{50}$ are equal.
Answer
The $(r+1)^{\text {th }}$ term of the expansion $(x+y)^n$ is given by $T_{r+1}={ }^n C_r x^{n-r} y^r$.
For the $17^{\text {th }}$ term, we have, $\mathrm{r}+1=17$, i.e., $\mathrm{r}=16$
Thus, $T_{17}=T_{16+1}={ }^{50} \mathrm{C}_{16}(2)^{50-16} a^{16}$
$={ }^{50} C_{16} 2^{34} a^{16}$
Similarly, $\mathrm{T}_{18}={ }^{50} \mathrm{C}_{17} 2^{33} \mathrm{a}^{17}$
Given that $T_{17}=T_{18}$
So, ${ }^{50} \mathrm{C}_{16}(2)^{34} a^{16}={ }^{50} \mathrm{C}_{17}(2)^{33} \mathrm{a}^{17}$
Thus,$\frac{^{50} C_{16} \cdot 2^{34}}{^{50} C_{17} \cdot 2^{33}}=\frac{a^{17}}{a^{16}}$
i.e., $a=\frac{^{50} \mathrm{C}_{16} \times 2}{^{50} \mathrm{C}_{17}}=\frac{50 !}{16 ! 34 !} \times \frac{17 ! \cdot 33 !}{50 !} \times 2=1$
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Question 61 Mark
Using binomial theorem, prove that $6^n–5n$ always leaves remainder $1$ when divided by $25$.
Answer
For two numbers a and b if we can find numbers q and r such that $a = bq + r$, then we say that b divides a with q as quotient and r as remainder. Therefore, in order to show that $n^6$– 5n leaves remainder 1 when divided by $25$, now we have to prove that $n^6– 5n = 25k + 1$, where k is some natural number.
Now,we have
$(1+a)^n={ }^n C_0+{ }^n C_1 a+{ }^n C_2 a^2+\ldots+{ }^n C_n a^n$
For $a=5$, we obtain
$(1+5)^n={ }^n C_0+{ }^n C_1 5+{ }^n C_2 5^2+\ldots+{ }^n C_n 5^n$
i.e. $(6)^n=1+5 n+5^2 \cdot{ }^n C_2+5^3 \cdot{ }^n C_3+\ldots+5^n$
i.e. $n 6^n-5 n=1+5^2\left({ }^n C_2+{ }^n C_3 5+\ldots+5^{n-2}\right)$
or $6^n-5 n=1+25\left({ }^n C_2+5 .{ }^n C_3+\ldots+5^{n-2}\right)$
or $6^n-5 n=25 k+1$ where $k={ }^n C_2+5 .{ }^n C_3+\ldots+5^{n-2}$.
Therefore, This shows that when divided by $25,6^n-5 n$ leaves remainder 1 .
Therefore,This shows that when divided by $25, 6^n – 5n$ leaves remainder 1.
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Question 71 Mark
Which is larger $(1.01)^{1000000}$ or $10,000 ?$
Answer
Splitting 1.01 and using binomial theorem to write the first few terms we have
$(1.1)^{10000}=(1+0.1)^{10000}$. Using binomial theorem
$={ }^{1000000} \mathrm{C}_0+{ }^{1000000} \mathrm{C}_1(0.01)+$ other positive terms
$=1+10000(0.01)+$ other positive terms
$=1+1000+$ other positive terms
$>10000$.
Therefore $(1.01)^{1000000}>10000$
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Question 81 Mark
Compute $(98)^5$.
Answer
Now,we express 98 as the sum or difference of two numbers whose powers are easier to calculate, and then use Binomial Theorem.
$\text { Write } 98=100-2$
$\text { Thus, }(98)^5=(100-2)^5$
$={ }^5 \mathrm{C}_0(100)^5-{ }^5 \mathrm{C}_1(100)^4 .2+{ }^5 \mathrm{C}_2(100)^3 2^2-{ }^5 \mathrm{C}_3(100)^2(2)^3+{ }^5 \mathrm{C}_4(100)(2)^4-{ }^5 \mathrm{C}_5(2)^5$
$=100000000000-5 \times 100000000 \times 2+10 \times 1000000 \times 4-10 \times 10000 \times 8+5 \times 100 \times 16-32$
$=10040008000-1000800032=9039207968$.
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Question 91 Mark
If the coefficients of $(r – 5)^{th}$ and $(2r – 1)^{th}$ terms in the expansion of $(1 + x)^{34}$ are equal, find r.
Answer
The coefficients of $(r-5)^{\text {th }}$ and $(2 r-1)^{\text {th }}$ terms of the expansion $(1+x)^{34}$ are ${ }^{34} C_{r-6}$ and ${ }^{34} C_{2 r-2}$, respectively. Since they are equal so ${ }^{34} \mathrm{C}_{\mathrm{r}-6}=$
$\mathrm{C}_{2 \mathrm{r}-2}$
Thus, either $\mathrm{r}-6=2 \mathrm{r}-2$ or $\mathrm{r}-6=34-(2 \mathrm{r}-2)$
[Using the fact that if ${ }^n C_r={ }^n C_p$, then either $r=p$ or $r=n-p$ ]
Therefore, we obtain , $r=-4$ or $r=14$. $r$ being a natural number, $r=-4$ is not possible.
Hence, $r=14$.
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Question 101 Mark
The sum of the coefficients of the first three terms in the expansion of $\left(x-\frac{3}{x^{2}}\right)^{m}, x \neq 0, m$ being a natural number, is $559$. Find the term of the expansion containing $x^3$.
Answer
The coefficients of the first three terms of $\left(x-\frac{3}{x^{2}}\right)^{m}$ are $^mC_0, (–3) ^mC_1$ and $9 ^mC_2$ . Thus,by the given condition, we have
$^{m} \mathrm{C}_{0}-3^{m} \mathrm{C}_{1}+9^{m} \mathrm{C}_{2}=559,$
i.e. $1-3 \mathrm{m}+\frac{9 \mathrm{m}(\mathrm{m}-1)}{2}=559$
which gives m = 12 (m being a natural number).
Now,we have $\mathrm{T}_{r+1}=^{12} \mathrm{C}_{r} x^{12-r}\left(-\frac{3}{x^{2}}\right)^{r}$
$=^{12} \mathrm{C}_{r}(-3)^{r} \cdot x^{12-3 r}$
Since we need the term containing $x^3$, Therefore put 12 – 3r = 3 i.e., r = 3.
Therefore, the required term is $^{12}C_3 (–3)^3 x^3$, i.e., $– 5940 x^3​​​​​​​$.
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Question 111 Mark
Find the term independent of x in the expansion of $\left(\sqrt[3]{x}+\frac{1}{2 \sqrt[3]{x}}\right)^{18}, x>0$.
Answer
We have ,$T_{r+1}$ = $=^{18} \mathrm{C}_{r}(\sqrt[3]{x})^{18-r}\left(\frac{1}{2 \sqrt[3]{x}}\right)^{r}$
$=^{18} \mathrm{C}_{r} x^{\frac{18-r}{3}} \cdot \frac{1}{2^{r} \cdot x^{\frac{r}{3}}}=^{18} \mathrm{C}_{r} \frac{1}{2^{r}} \cdot x^{\frac{18-2 r}{3}}$
Now,we have to find a term independent of x, i.e., term not having x, so take $\frac{18-2 r}{3}=0$.
We obtain r = 9. The required term is $^{\mathrm{18}} \mathrm{C}_{9} \frac{1}{2^{9}}$.
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Question 121 Mark
Find the $r^{\text {th }}$ term from the end in the expansion of $(x+a)^n$
Answer
We have to given there are $(n+1)$ terms in the expansion of $(x+a)^n$.
Observing the terms we can say that the first term from the end is the last term, i.e., $(n+1)^{\text {th }}$ term of the expansion and $n+1=(n+1)-(1-1)$.
The second term from the end is the $n^{\text {th }}$ term of the expansion, and $n=(n+1)-(2-1)$.
The third term from the end is the $(n-1)$ term of the expansion and $n-1=(n+1)-(3-1)$ and so on.
Therefore, rth term from the end will be term number $(n+1)-(r-1)=(n-r+2)$ of the expansion.
Therefore, the $(n-r+2)^{\text {th }}$ term is ${ }^n C_{n-r+1} x^{r-1} a^{n-r+1}$.
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Question 131 Mark
Find the coefficient of $a^4$ in the product $(1 + 2a)^4 (2 – a)^5$ using the binomial theorem.
Answer
We first expand each of the factors of the given product using Binomial Theorem. Here, we have
$\Longrightarrow(1+2 a)^4={ }^4 C_0+{ }^4 C_1(2 a)+{ }^4 C_2(2 a)^2+{ }^4 C_3(2 a)^3+{ }^4 C_4(2 a)^4$
$=1+4(2 a)+6\left(4 a^2\right)+4\left(8 a^3\right)+16 a^4$
$=1+8 a+24 a^2+32 a^3+16 a^4$
and also $(2-\mathrm{a})^5={ }^3 \mathrm{C}_0(2)^5-{ }^5 \mathrm{C}_1(2)^4(a)+{ }^5 \mathrm{C}_2(2)^3(a)^2-{ }^5 \mathrm{C}_3(2)^2(a)^3+{ }^5 \mathrm{C}_4(2)(a)^4-{ }^5 \mathrm{C}_5(a)^5$
$=32-80 a+80 a^2-40 a^3+10 a^4-a^5$
Therefore, $(1+2 a)^4(2-a)^5=\left(1+8 a+24 a^2+32 a^3+16 a^4\right)\left(32-80 a+80 a^2-40 a^3+10 a^4-a^5\right)$
The complete multiplication of the two brackets need not be carried out.
We write only those terms which involve $a^4$
This can be done if we note that $a^r \cdot a^{4-r}=a^4$. The terms containing $a^4$ are
$1\left(10 a^4\right)+(8 a)\left(-40 a^3\right)+\left(24 a^2\right)\left(80 a^2\right)+\left(32 a^3\right)(-80 a)+\left(16 a^4\right)(32)=-438 a^4$
Therefore, the required coefficient of $\mathrm{a}^4$ in the given product is -438
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Question 141 Mark
Show that the coefficient of the middle term in the expansion of $(1 + x)^{2n}$ is equal to the sum of the coefficients of middle terms in the expansion of $(1 + x)^{2n-1}$.
Answer
As discussed in the previous example, the middle term in the expansion of $(1 + x)^{2n}$ is given by $T_{n+1}=\;^{2 n} C_{n} x^{n}$
So, the coefficient of the middle term in the expansion of $(1 + x)^{2n}​​​​​​​$ is $^{2 n} \mathrm{C}_{n}$.
Now, consider the expansion of $(1 + x)^{2n-1}​​​​​​​$ Here, the index (2n-1) is odd.
So, $\left(\frac{(2 n-1)+1}{2}\right)^{t h}$ and $\left(\frac{(2 n-1)+1}{2}+1\right)^{\text { th }}$i.e., $n^{th}$ and $(n + 1)^{th}$ terms are middle terms.
Now, $T_{n}=T_{(n-1)+1}$,= $^{2 n-1} C_{n-1}(1)^{(2 n-1)-(n-1)} x^{n-1}$ = $^{2 n-1} C_{n-1} x^{n-1}$
and, $T_{n+1} = ^{2 n-1} C_{n}(1)^{(2 n-1)-n} x^{n}=\; ^{2 n-1}C_{n} x^{n}$
So, the coefficients of two middle terms in the expansion of $(1 + x)^{2n-1}​​​​​​​$ are $^{2 n-1} C_{n-1}$ and $^{2n-1}{C_{n}}$.
$\therefore$ Sum of these coefficients = $^{2 n-1} C_{n-1}+^{2 n-1} C_{n}$
= $^{(2 n-1)+1}{C_{n}}$ [$\because$ $^{n} C_{r-1}+^{n} C_{r}=^{n+1} C_{r}$]
= $^{2 n} C_{n}$
= Coefficient of middle term in the expansion of $(1 + x)^{2n}​​​​​​​$.
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Question 151 Mark
If the coefficients of $a^{r-1}, a^r$ and $a^{r+1}$ in the expansion of $(1+a)^n$ are in arithmetic progression, prove that $n^2-n(4 r+$ 1) $+4 r^2-2=0$.
Answer
We have the $(r+1)^{\text {th }}$ term in the expansion is ${ }^n C_r a^r$. Therefore, it can be seen that $a^r$ occurs in the $(r+1)^{\text {th }}$ term, and its coefficient is ${ }^n \mathrm{Cr}$. Therefore, the coefficients of $a^{r-1}$, $a^r$ and $a^{r+1}$ are ${ }^n C_{r-1},{ }^n C_r$ and ${ }^n C_{r+1}$, respectively. Since these coefficients are in arithmetic progression, so we have, ${ }^n C_{r-1}+{ }^n C_{r+1}=2 .^n \mathrm{Cr}$ it is given
$\frac{n !}{(r-1) !(n-r+1) !}+\frac{n !}{(r+1) !(n-r-1) !}=2 \times \frac{n !}{r !(n-r) !}$
i.e. $\frac{1}{(r-1) !(n-r+1)(n-r)(n-r-1) !}+\frac{1}{(r+1)(r)(r-1) !(n-r-1) !}$ $=2 \times \frac{1}{r(r-1) !(n-r)(n-r-1) !}$
or $\frac{1}{(r-1) !(n-r-1) !}\left[\frac{1}{(n-r)(n-r+1)}+\frac{1}{(r+1)(r)}\right]$ $=2 \times \frac{1}{(r-1) !(n-r-1) ![r(n-r)]}$
i.e. $\frac{1}{(n-r+1)(n-r)}+\frac{1}{r(r+1)}=\frac{2}{r(n-r)}$,
or $\frac{r(r+1)+(n-r)(n-r+1)}{(n-r)(n-r+1) r(r+1)}=\frac{2}{r(n-r)}$
$\text { or } r(r+1)+(n-r)(n-r+1)=2(r+1)(n-r+1)$
$\text { or } r^2+r+n^2-n r+n-n r+r^2-r=2\left(n r-r^2+r+n-r+1\right)$
$\text { or } n^2-4 n r-n+4 r^2-2=0$
$\text { so, } n^2-n(4 r+1)+4 r^2-2=0$
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Question 161 Mark
Find the term independent of x in the expansion of $\left( \frac { 3 } { 2 } x ^ { 2 } - \frac { 1 } { 3 x } \right) ^ { 6 }$.
Answer
General term in the expansion of $\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^6$ is
$T_{r+1}={}^6 \mathrm{C}_r\left(\frac{3}{2} x^2\right)^{6-r}\left[\frac{-1}{3 x}\right]^r$
$={ }^6 \mathrm{C}_r\left(\frac{3}{2}\right)^{6-r} \times\left(\mathrm{x}^2\right)^{6-\mathrm{r}}(-1)^r\left(\frac{1}{3^r}\right)\left(\frac{1}{x}\right)^r$
$=(-1)^{r 6} \mathrm{C}_r \frac{(3)^{6-r}}{(3)^r \cdot(2)^{6-r}} \times x^{12-2 r} \times \frac{1}{x^r}$
$=(-1)^{r 6} C_r \frac{(3)^{6-2 r}}{(2)^{6-r}} \times x^{12-3 r}$
Now, the term will be independent of x , if the index of x is zero, i.e.,
$12-3 r=0 \Rightarrow r=4$
Hence, 5 th term is independent of x and is given by
$T_5=(-1)^{46} C_r \frac{(3)^{6-8}}{(2)^{6-4}}=\frac{5}{12}$
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Question 171 Mark
Expand $\left( x ^ { 2 } + \frac { 3 } { x } \right) ^ { 4 }$, x $\neq$ 0
Answer
By using binomial theorem, we have
$(a+b)^n={ }^n C_0 a^n+{ }^n C_1 a^{n-1} b+{ }^n C_2 a^{n-2} b^2+\ldots+{ }^n C_r a^{n-r} b^r+\ldots+{ }^n C_n b^n$
$\therefore\left(x^2+\frac{3}{x}\right)^4={ }^4 C_0\left(x^2\right)^4+{ }^4 C_1\left(x^2\right)^{4-1}\left(\frac{3}{x}\right)^1+{ }^4 C_2\left(x^2\right)^{4-2}\left(\frac{3}{x}\right)^2+{ }^4 C_3\left(x^2\right)^{4-3}\left(\frac{3}{x}\right)^3+4 C_4\left(x^2\right)^{4-4}\left(\frac{3}{x}\right)^4$
$=1 \cdot x^8+4\left(x^2\right)^3\left(\frac{3}{x}\right)+6\left(x^2\right)^2\left(\frac{3}{x}\right)^2+4\left(x^2\right)\left(\frac{3}{x}\right)^3+1 \cdot\left(x^2\right)^0\left(\frac{3}{x}\right)^4$
${\left[\because{ }^4 C_0=1,{ }^4 C_1=4,{ }^4 C_2=6,{ }^4 C_3=4 \text { and }{ }^4 C_4=1\right]}$
$=x^8+4 \cdot x^6 \cdot \frac{3}{x}+6 x^4 \cdot \frac{9}{x^2}+4 x^2 \cdot \frac{27}{x^3}+1 \cdot 1 \cdot \frac{81}{x^4}$
$=x^8+12 x^5+54 x^2+\frac{108}{x}+\frac{81}{x^4}$
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