Question 15 Marks
If the $2^{nd}, 3^{rd}$ and $4^{th}$ terms in the expansion of $(\text{x}+\text{a})^{\text{n}}$ are $240, 729$ and $1080$ respectively find $x, a, n.$
AnswerIf the the expansion of $(\text{x}+\text{a})^{\text{n}},$ the $2^{nd}, 3^{rd}$ and $4^{th}$ are ${^\text{n}}\text{C}_{\text{1}}\text{x}^{\text{n}-1}\text{a}^{1}, \ {^\text{n}}\text{C}_{\text{2}}\text{x}^{\text{n}-2}\text{a}^{2}$ and,
According to the quation,
${^\text{n}}\text{C}_{\text{1}}\text{x}^{\text{n}-1}\text{a}^{1}=240$
${^\text{n}}\text{C}_{\text{2}}\text{x}^{\text{n}-2}\text{a}^{2}=720$
${^\text{n}}\text{C}_{\text{3}}\text{x}^{\text{n}-3}\text{a}^{3}=1080$
Now,
$\Rightarrow \frac{{^\text{n}}\text{C}_{\text{2}}\text{x}^{\text{n}-2}\text{a}^{2}}{{^\text{n}}\text{C}_{\text{3}}\text{x}^{\text{n}-3}\text{a}^{3}}=\frac{720}{240}$
$\Rightarrow \frac{\text{n}-1}{2\text{x}}\text{a}=3$
$\Rightarrow \frac{\text{a}}{\text{x}}=\frac{6}{\text{n}...1}\ ...(\text{i})$
Also,
$\frac{{^\text{n}}\text{C}_{\text{3}}\text{x}^{\text{n}-3}\text{a}^{3}}{{^\text{n}}\text{C}_{\text{2}}\text{x}^{\text{n}-2}\text{a}^{2}}=\frac{1080}{720}$
$\Rightarrow \frac{\text{n}-2}{3\text{x}}\text{a}=\frac{3}{2}$
$\Rightarrow \frac{\text{a}}{\text{x}}=\frac{9}{2\text{n}...4}\ ...(\text{ii})$
From (i) and (ii), we get
$\frac{6}{\text{n}-1}=\frac{9}{2\text{n}-4}$
$\Rightarrow \text{n}=5$
Putting in equation (i), we get
$\Rightarrow 2\text{a}=3\text{x}$
Now,
${^\text{5}}\text{C}_{\text{1}}\text{x}^{\text{5}-1}\big(\frac{\text{3}}{2}\text{x}\big)=240$
$\Rightarrow 15\text{x}^{5}=480$
$\Rightarrow \text{x}^{5}=32$
$\Rightarrow \text{x}=2$
By putting the value of $x$ and $n$ in $(i).$ we get
$\text{a}=3$
View full question & answer→Question 25 Marks
Evaluate the following:
$\Big\{\text{a}^2+\sqrt{\text{a}^2-1}\Big\}^4+\Big\{\text{a}^2-\sqrt{\text{a}^2-1}\Big\}^4$
Answer$\Big\{\text{a}^2+\sqrt{\text{a}^2-1}\Big\}^4+\Big\{\text{a}^2-\sqrt{\text{a}^2-1}\Big\}^4$
$\text{Let}\ \text{a}^2=\text{A},\ \sqrt{\text{a}^2-1}=\text{B}$
$(\text{A}+\text{B})^4+(\text{A}-\text{B})^4$
$=\text{B}^4+{^4\text{C}}_1\text{AB}^3+{^4\text{C}}_2\text{A}^2\text{B}^2+{^4\text{C}}_3\text{A}^3\text{B}+\text{A}^2+\text{B}^4\\-{^4\text{C}}_1\text{AB}^3+{^4\text{C}}_2\text{A}^2\text{B}^2-{^4\text{C}}_3\text{A}^3\text{B}+\text{A}^4$
$=2\big(\text{A}^4+{^4\text{C}}_2\text{A}^2\text{B}^2+\text{B}^4\big)$
$=2\big(\text{A}^2+6\text{A}^2\text{B}^2+\text{B}^4\big)$
$=2\Big(\text{a}^8+6\text{a}^4(\text{a}^2-1)+(\text{a}^2-1)^2\Big)$
$=2\Big[\text{a}^8+6\text{a}^6-6\text{a}^4+\text{a}^4+1-2\text{a}^2\Big]$
$\Big\{\text{a}^2+\sqrt{\text{a}^2-1}\Big\}+\Big\{\text{a}^2-\sqrt{\text{a}^2-1}\Big\}$
$=2\text{a}^8+12\text{a}^6-10\text{a}^4-4\text{a}^4+2$
View full question & answer→Question 35 Marks
If n is a positive integer, prove that $3^{3\text{n}}-26\text{n}-1$ is divisible by 676.
Answer$3^{3\text{n}}-26\text{n}-1$
$=(3^3)^\text{n}-26\text{n}-1$
$=27^\text{n}-26\text{n}-1$
$=(1+26)^\text{n}-26\text{n}-1$
$\Big({^\text{n}\text{C}}_0+{^\text{n}\text{C}}_1(26)^1+{^\text{n}\text{C}}_2(26)^2+.....+676(26)^{\text{n}-2}\Big)-26\text{n}-1$
$=676\Big({^\text{n}\text{C}}_2+......+(26)^{\text{n}-2}\Big)$
$\therefore3^{3\text{n}}-26\text{n}-1$ is divisible for $\text{n}\in\text{N}.$
Hence, proved
View full question & answer→Question 45 Marks
Find the term independent of $x$ in the expansion of the following expressions:
$(1+\text{x}+2\text{x}^{3})\Big(\frac{3}{2}\text{x}^{2}-\frac{1}{3\text{x}}\Big)^{9}$
Answer$(1+\text{x}+2\text{x}^{3})\Big(\frac{3}{2}\text{x}^{2}-\frac{1}{3\text{x}}\Big)^{9}$
$(1+\text{x}+2\text{x}^{3})\bigg[\Big(\frac{3}{2}\text{x}^{2}\Big)^{9}-{^\text{9}}\text{C}_{\text{1}}\Big(\frac{3}{2}\text{x}^{2}\Big)^{8}\ \frac{1}{3\text{x}}.....\\ +{^\text{9}}\text{C}_{\text{6}}\Big(\frac{3}{2}\text{x}^{2}\Big)^{3}\Big(\frac{1}{3\text{x}}\Big)^{6}-{^\text{9}}\text{C}_{\text{7}}\Big(\frac{3}{2}\text{x}^{2}\Big)^{2}\Big(\frac{1}{3\text{x}}\Big)^{7}\bigg]$
In the second bracket, we have to search the term so $x^0$ and $\frac{1}{\text{x}^{3}}$ which when multiplying by $1$ and $2x^3$ is first bracket will give the term independent of $x.$ The term containing $\frac{1}{\text{x}}$ will not occur is second bracket.
The teram independent of $x,$
$=1\Big[{^\text{9}}\text{C}_{\text{6}}\frac{3^{3}}{2^{2}}\times\frac{1}{3^{6}}\Big]-2\text{x}^{3}\Big[{^\text{9}}\text{C}_{\text{7}}\frac{3^{3}}{2^{2}}\times\frac{1}{3^{7}}\times\frac{1}{\text{x}^{3}}\Big]$
$=\Big[\frac{9\times8\times7}{1\times2\times3}\times\frac{1}{8\times27}\Big]-2\Big[\frac{9\times8}{1\times2}-\frac{1}{4\times243}\Big]$
$=\frac{7}{18}-\frac{2}{27}$
$=\frac{17}{54}$
View full question & answer→Question 55 Marks
Find the sixth term in the expansion $\Big(\text{y}^{\frac{1}{2}}+\text{x}^{\frac{1}{3}}\Big)^{\text{n}},$ if the binomial coefficient of the term from the end is $45.$
AnswerIn the binomial expansion of $\Big(\text{y}^{\frac{1}{2}}+\text{x}^{\frac{1}{3}}\Big)^{\text{n}},$ there are $(n + 1)$ terms.
The third term from the end in the expansion of $\Big(\text{y}^{\frac{1}{2}}+\text{x}^{\frac{1}{3}}\Big)^{\text{n}},$ is the third term from the beginning in the expansion of $\Big(\text{x}^{\frac{1}{3}}+\text{y}^{\frac{1}{2}}\Big)^{\text{n}}.$
The binomial coefficient of the third term from the end is $45.$
$\therefore {^\text{n}}\text{C}_{\text{2}}=45$
$\Rightarrow \frac{\text{n}(\text{n}-1)}{2}=45$
$\Rightarrow \text{n}^{2}-\text{n}-90=0$
$\Rightarrow (\text{n}-10)(\text{n}+9)=0$
$\Rightarrow \text{n}=10$
Let $T_6$ be the $six^{th}$ term in the binomoal expansion of $\Big(\text{y}^{\frac{1}{2}}+\text{x}^{\frac{1}{3}}\Big)^{\text{n}}.$
Then,
$\text{T}_{6}={^\text{n}}\text{C}_{\text{5}}\Big(\text{y}^{\frac{1}{2}}\Big)^{\text{n}-5}\Big(\text{x}^{\frac{1}{3}}\Big)^{5}$
$={^\text{10}}\text{C}_{\text{5}}\text{y}^{\frac{5}{2}}\text{x}^{\frac{5}{3}}$
$=252\text{y}^{\frac{5}{2}}\text{x}^{\frac{5}{3}}$
Hence, the $six^{th}$ term in the expansion of $\Big(\text{y}^{\frac{1}{2}}+\text{x}^{\frac{1}{3}}\Big)^{\text{n}},$ is $252\text{y}^{\frac{5}{2}}\text{x}^{\frac{5}{3}}.$
View full question & answer→Question 65 Marks
Using binomial evaluate the following:
$(98)^5$
AnswerWe have,
$(98)^5=(100-2)^5$
$={^5\text{C}}_0\times100^5+{^5\text{C}}_1\times100^4\times(-2)+{^5\text{C}}_2\times100^3\times(-2)^2\\+{^5\text{C}}_3\times100^2\times(-2)^3+{^5\text{C}}_4\times100\times(-2)^4+{^5\text{C}}_5\times(-2)^5$
$={^5\text{C}}_0\times100^5-{^5\text{C}}_1\times100^4\times2+{^5\text{C}}_2\times100^3\times4\\-{^5\text{C}}_3\times100^2\times8+{^5\text{C}}_4\times100\times16-{^5\text{C}}_5\times32$
$=100^5-10\times100^4+40\times100^3-80\times100^2+80\times100-32$
$=10000000000-1000000000+40000000-800000+8000-32$
$=10040008000-1000800032$
$=9039207968$
$\therefore(98)^5=9039207968$
View full question & answer→Question 75 Marks
Find the middle terms(s) in the expansion of:
$\Big(2\text{x}-\frac{\text{x}^{2}}{4}\Big)^{9}$
Answer$\Big(2\text{x}-\frac{\text{x}^{2}}{4}\Big)^{9}$
Here, n is an odd number.
Therefore, the middle terms are $\Big(\frac{\text{n+1}}{2}\Big)$ and $\Big(\frac{\text{n+1}}{2}+1\Big)$ 5th and 6th terms.
Now, we have,
$\text{T}_{5}=\text{T}_{4+1}$
$={^\text{9}}\text{C}_{\text{4}}(2\text{x})^{9-4}\big(\frac{-\text{x}^{2}}{4}\big)^{4}$
$=\frac{9\times8\times7\times6}{4\times3\times2}\times2^{5}\frac{1}{4^{4}}\text{x}^{5+8}$
$=\frac{63}{4}\text{x}^{13}$
View full question & answer→Question 85 Marks
If $a, b, c$ and $d$ in any binomial expansion be the $6^{th}, 7^{th}, 8^{th}$ and $9^{th}$ terms respectively, then prove that $\frac{\text{b}^{2}-\text{ac}}{\text{c}^{2}-\text{bd}}=\frac{4\text{a}}{3\text{c}}.$
Answer$T_6 = a$
$T_7 = b$
$T_8 = c$
$T_9 = d$
We have, $(x + 1)^n$
Then,
$\frac{\text{T}_{\text{r}+1}}{\text{T}_{\text{r}}}=\frac{\text{n-r}+1}{\text{r}}.\frac{1}{\text{x}}$
Put $r = 6,$
$\frac{\text{T}_7}{\text{T}_6}=\frac{\text{n}-6+1}{6}.\frac{1}{\text{x}}=\frac{\text{n}-5}{6}.\frac{1}{\text{x}}=\frac{\text{b}}{\text{a}}\ \dots(\text{a})$
Put $r = 7$
$\frac{\text{T}_8}{\text{T}_7}=\frac{\text{n}-7+1}{7}.\frac{1}{\text{x}}=\frac{\text{n}-6}{7}.\frac{1}{\text{x}}=\frac{\text{c}}{\text{b}}\ \dots(\text{b})$
Put $r = 8$
$\frac{\text{T}_9}{\text{T}_8}=\frac{\text{n}-8+1}{8}.\frac{1}{\text{x}}=\frac{\text{n}-7}{7}.\frac{1}{\text{x}}=\frac{\text{d}}{\text{c}}\ \dots(\text{c})$
Now,
$\frac{\text{b}^{2}-\text{ac}}{\text{c}^{2}-\text{bd}}=\frac{4\text{a}}{3\text{c}}$
Divide by bc in L.H.S, we get
$\Rightarrow \frac{\frac{\text{b}^{2}}{\text{bc}}-\frac{\text{ac}}{\text{bc}}}{\frac{\text{c}^{2}}{\text{bc}}-\frac{\text{bd}}{\text{bc}}}=\frac{4\text{a}}{3\text{c}}$
$\Rightarrow \frac{\frac{\text{b}}{\text{c}}-\frac{\text{a}}{\text{b}}}{\frac{\text{c}}{\text{b}}-\frac{\text{d}}{\text{c}}}=\frac{4\text{a}}{3\text{c}}$
Take L.H.S
By Equation $(a), (b), (c)$
$\Rightarrow\ \frac{\frac{7\text{x}}{\text{n}-6}-\frac{6\text{x}}{\text{n}-5}}{\frac{\text{n}-6}{7\text{x}}-\frac{\text{n}-7}{8\text{x}}}$
$\Rightarrow\ \frac{\text{x}\Big[\frac{7}{\text{n}-6}-\frac{6}{\text{n}-5}\Big]}{\frac{1}{\text{x}}\Big[\frac{\text{n}-6}{7}-\frac{\text{n}-7}{8}\Big]}$
$\Rightarrow\ \text{x}^2\Bigg[\frac{\text{n}-35-6\text{n}+36}{\big(\text{n}-6\big)\big(\text{n}-5\big)}\Bigg]\times\frac{56}{8\text{n}-48-7\text{n}+49}$
On solving this, we get
$\frac{56\text{x}^2}{\big(\text{n}-6\big)\big(\text{n}-5\big)}\ \dots(\text{X})$
On Taking R.H.S
$\Rightarrow\ \frac{4\text{a}}{3\text{c}}=\frac{4}{3}\Big(\frac{\text{a}}{\text{b}}\times\frac{\text{b}}{\text{c}}\Big)$
On putting the values from equation $(a), (b), (c)$
$\Rightarrow\ \frac{4}{3}\Big[\frac{6\text{x}}{\text{n}-5}\times\frac{7\text{x}}{\text{n}-6}\Big]$
$\frac{56\text{x}^2}{\big(\text{n}-6\big)\big(\text{n}-5\big)}\ \dots(\text{Y})$
By equation $X$ and $Y,$ it is proved that
L.H.S = R.H.S
View full question & answer→Question 95 Marks
Find the coefficients of $a^4$ in the product $(1+2\text{a})^{4}(2-\text{a})^{5}$ using binomial theorem.
AnswerWe have,
$(1+2\text{a})^{4}(2-\text{a})^{5}$
$=\Big[{^\text{4}}\text{C}_{\text{0}}\big(2\text{a}\big)^{0}+{^\text{4}}\text{C}_{\text{1}}\big(2\text{a}\big)^1+{^\text{4}}\text{C}_{\text{2}}\big(2\text{a}\big)^{2}+{^\text{4}}\text{C}_{\text{3}}\big(2\text{a}\big)^{3}+{^\text{4}}\text{C}_{\text{4}}\big(2\text{a}\big)^{4}\Big]\\ \times\Big[{^\text{5}}\text{C}_{\text{0}}(2)^{5}(-\text{a})^{0}+{^\text{5}}\text{C}_{\text{1}}(2)^{4}(-\text{a})^{1}+{^\text{5}}\text{C}_{\text{2}}(2)^{3}(-\text{a})^{2}+{^\text{5}}\text{C}_{\text{2}}(2)^{2}(-\text{a})^{3}+{^\text{5}}\text{C}_{\text{1}}(2)^{1}(-\text{a})^{5}$
$=\Big[1+8\text{a}+24\text{a}^{2}+32\text{a}^{3}+16\text{a}^{4}\Big]\times\Big[32-80\text{a}+80\text{a}^{2}-40\text{a}^{3}+10\text{a}^{4}-\text{a}^{5}\Big]$
Coefficient of $\text{a}^{4}=10-320+1920-2560+512=-438$
View full question & answer→Question 105 Marks
Using binomial evaluate the following:
$(102)^5$
AnswerWe have,
$(102)^5=(100+2)^5$
$={^5\text{C}}_0\times100^5+{^5\text{C}}_1\times100^4\times2+{^5\text{C}}_2\times100^3\times2^2\\+{^5\text{C}}_3\times100^2\times2^3+{^5\text{C}}_4\times100\times2^4+{^5\text{C}}_5\times2^5$
$=100^5+5\times100^4\times2+10\times100^3\times2^2\\+10\times100^2\times2^3+5\times100\times2^4+2^5$
$=10000000000+1000000000+40000000+800000+8000+32$
$=11040808032$
$\therefore(102)^5=11040808032$
View full question & answer→Question 115 Marks
If the coefficients of three consecutive terms in the expansion of $(1+\text{x})^{\text{n}}$ be 76, 95 and 76, find n.
AnswerWe have,
$(1+\text{x})^{\text{n}}$
Suppose r, (r + 1) and (r + 2) are three consecutive terms in the given expansion.
The coefficirnts of these terms are ${^\text{n}}\text{C}_{\text{r}-1},{^\text{n}}\text{C}_{\text{r}}, {^\text{n}}\text{C}_{\text{r}+1}.$
Coefficients of rth terms $={^\text{n}}\text{C}_{\text{r}-1}=76$
Coefficients of (r + 1) terms $={^\text{n}}\text{C}_{\text{r}}=95$
Coefficients of (r + 2) terms $={^\text{n}}\text{C}_{\text{r}-1}=76$
$\Rightarrow {^\text{n}}\text{C}_{\text{r}-1}={^\text{n}}\text{C}_{\text{r}+1}$
$\Rightarrow \text{r}-1+\text{r}+1=\text{n}$
$\Rightarrow \text{r}=\frac{\text{n}}{2}$
Now,
$\therefore \frac{{^\text{n}}\text{C}_{\text{r}}}{{^\text{n}}\text{C}_{\text{r}-1}}=\frac{96}{76}$
$\Rightarrow \frac{\text{n}-\text{r}+1}{\text{r}}=\frac{95}{76}$
$\Rightarrow \frac{\frac{\text{n}}{2}+1}{\frac{\text{n}}{2}}=\frac{95}{76}$
$\Rightarrow 38\text{n}+76=\frac{95\text{n}}{2}$
$\Rightarrow \frac{19\text{n}}{2}=76$
$\Rightarrow \text{n}=8$
View full question & answer→Question 125 Marks
If in the expansion of $(1+\text{x})^{\text{n}},$ the coefficients of pth and qth term are equal, prove that $\text{p}+\text{q}=\text{n}+2,$ where $\text{p}\neq\text{q}.$
AnswerWe have,
$(1+\text{x})^{\text{n}}$
Coefficient of pth term $={^\text{n}}\text{C}_{\text{p}-1}$
Coefficient of qth term $={^\text{n}}\text{C}_{\text{q}-1}$
It is given that, these coefficients are equal.
${^\text{n}}\text{C}_{\text{p}-1}={^\text{n}}\text{C}_{\text{q}-1}$
$\Rightarrow \text{p}-1=\text{q}-1$ or $\text{p}-1+\text{q}-1=\text{n}$
$\Rightarrow \text{p}-\text{q}=0$ or $\text{p}+\text{q}=\text{n}+2$
$\therefore\ \text{p}+\text{q}=\text{n}+2$
Hence proved.
View full question & answer→Question 135 Marks
Show that $2^{4\text{n}+4}-15\text{n}-16,$ where $\text{n}\in\text{N}$ is divisible by 225.
Answer$2^{4\text{n}+4}- 15\text{n}-16=2^{4(\text{n}+1)}-15\text{n}-15-1$
$=(16)^{\text{n}+1}-15(\text{n}+1)-1$
$=(1+15)^{\text{n}+1}-15(\text{n}+1)-1$
$=\big[{^{\text{n}+1}\text{C}}_0+{^{\text{n}+1}\text{C}}_1(15)+{^{\text{n}+1}\text{C}}_2(15)^2+.....\\+{^{\text{n}+1}\text{C}}_{\text{n}+1}(15)^{\text{n}+1}\big]-15(\text{n}+1)-1$
$=\big[1+15({\text{n}+1})+{^{\text{n}+1}\text{C}}_2(15)^2+.....\\+{^{\text{n}+1}\text{C}}_{\text{n}+1}(15)^{\text{n}+1}\big]-15(\text{n}+1)-1$
$=225\big[{^{\text{n}+1}\text{C}}_2+.....{^{\text{n}+1}\text{C}}_{\text{n}+1}(15)^{\text{n}-1}\big]$
= 225 × natural number
View full question & answer→Question 145 Marks
If the seventh term from the beginning and in the binomial expansion of $\Big(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\Big)^{\text{n}}$ are equal, is the $7^{th}$ term from the end.
AnswerIn the binomial expansion of $\Big(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\Big)^{\text{n}},\big[(\text{n}+1)-7+1\big]^\text{th}\text{i.e.,}\ (\text{n}-5)^{\text{th}}$ term from the beginning is the $7^{th}$ term from the end.
Now,
$\text{T}_{7}={^\text{n}}\text{C}_{\text{6}}\Big(\sqrt[3]{2}\Big)^{\text{n}-6}\Big(\frac{1}{\sqrt[3]{3}}\Big)^{6}={^\text{n}}\text{C}_{\text{6}}\times2^{\frac{\text{n}}{3}-2}\times\frac{1}{3^{2}}$
And,
$\text{T}_{\text{n}-5}={^\text{n}}\text{C}_{\text{n}-6}\Big(\sqrt[3]{2}\Big)^{6}\Big(\frac{1}{\sqrt[3]{3}}\Big)^{\text{n}-6}={^\text{n}}\text{C}_{\text{6}}\times2^{2}\times\frac{1}{3^{\frac{\text{n}}{3}-2}}$
It is given that,
$\text{T}_{7}=\text{T}_{\text{n}-5}$
$\Rightarrow {^\text{n}}\text{C}_{\text{6}}\times2^{\frac{\text{n}}{3}-2}\times\frac{1}{3^{2}}$
$\Rightarrow {^\text{n}}\text{C}_{\text{6}}\times2^{2}\times\frac{1}{3^{\frac{\text{n}}{3}-2}}$
$\Rightarrow \frac{2^{\frac{\text{n}}{3}-2}}{2^{2}}=\frac{3^{2}}{3^{\frac{\text{n}}{3}-2}}$
$\Rightarrow (6)^{\frac{\text{n}}{3}-2}=6^{2}$
$\Rightarrow \frac{\text{n}}{3}-2=2$
$\Rightarrow \text{n}=12$
Hence, the value of $n$ is $12.$
View full question & answer→Question 155 Marks
Find the term independent of $x$ in the expansion of the following expressions:
$\Big(\frac{3}{2}\text{x}^{2}-\frac{1}{3\text{x}}\Big)^{6}$
AnswerWe have,
$\Big(\frac{3}{2}\text{x}^{2}-\frac{1}{3\text{x}}\Big)^{6}$
In expansion,
$\text{T}_{\text{r+1}}= {^\text{6}}\text{C}_{\text{r}}\Big(\frac{3\text{x}^{2}}{2}\Big)^{6-\text{r}}\Big(-\frac{1}{3\text{x}}\Big)^{\text{r}}$
$= {^\text{6}}\text{C}_{\text{r}}\Big(\frac{3}{2}\Big)^{6-\text{r}}(\text{x}^{12-3\text{r}})\Big(-\frac{1}{3}\Big)^{\text{r}}$
Let $T_{r+1}$ be independent of $x,$
$12-3\text{r}=0$
$\text{r}=4$
Required term
$\text{T}_{\text{r+1}}= \text{T}_{\text{4+1}}=\text{T}_{\text{5}}={^\text{6}}\text{C}_{\text{r}}\Big(\frac{3}{2}\Big)^{6-\text{4}}\Big(\frac{-1}{3}\Big)^{4}\text{x}^{12-3(4)}$
$=15\Big(\frac{9}{4}\Big)\Big(\frac{1}{81}\Big)\text{x}^{0}$
$=\frac{5}{12}$
View full question & answer→Question 165 Marks
Find the term independent of $x$ in the expansion of the following expressions:
$\Big(\frac{1}{2}\text{x}^{\frac{1}{3}}+\text{x}^{-\frac{1}{5}}\Big)^{8}$
AnswerWe have,
$\Big(\frac{1}{2}\text{x}^{\frac{1}{3}}+\text{x}^{-\frac{1}{5}}\Big)^{8}$
Let $(r + 1)^{th}$ term be independent of $x.$
$\text{T}_{\text{r}+1}={^\text{8}}\text{C}_{\text{r}}\Big(\frac{1}{2}\text{x}^{\frac{1}{3}}\Big)^{8-\text{r}}\Big(\text{x}^{\frac{-1}{5}}\Big)^{8}$
$={^\text{8}}\text{C}_{\text{r}}\Big(\frac{1}{2}\Big)^{8-\text{r}}\times\Big(\text{x}^\frac{1}{3}\Big)^{8-\text{r}}\times\Bigg(\frac{1}{\text{x}^{\frac{1}{5}}}\Bigg)^{\text{r}}$
$={^\text{8}}\text{C}_{\text{r}}\Big(\frac{1}{2}\Big)^{8-\text{r}}\times\big(\text{x}\big)^\frac{8-\text{r}}{3}\times\Bigg(\frac{1}{\text{x}^{\frac{1}{5}}}\Bigg)$
$={^\text{8}}\text{C}_{\text{r}}\Big(\frac{1}{2}\Big)^{8-\text{r}}\times\big(\text{x})^{\frac{8-\text{r}}{3}-\frac{\text{r}}{5}}$
$={^\text{8}}\text{C}_{\text{r}}\Big(\frac{1}{2}\Big)^{8-\text{r}}\times\big(\text{x})^{\frac{40-5\text{r}-3\text{r}}{15}}$
$={^\text{8}}\text{C}_{\text{r}}\Big(\frac{1}{2}\Big)^{8-\text{r}}\times\big(\text{x})^{\frac{40-8\text{r}}{15}}$
If it is independent of $x,$ We must have
$\Rightarrow \frac{40-8\text{r}}{15}=0$
$\Rightarrow 8\text{r}=40$
$\Rightarrow \text{r}=5$
The term independet of $x,$
Now,
$\text{T}_{6}={^\text{8}}\text{C}_{\text{r}}\Big(\frac{1}{2}\text{x}^{\frac{1}{3}}\Big)^{8-5}\Big(\text{x}^{\frac{-1}{5}}\Big)^{5}$
$=56\times\Big(\frac{1}{2}\Big)^{3}$
$=56\times\frac{1}{8}$
$=7$
View full question & answer→Question 175 Marks
In the expansion of $(1+\text{x})^{\text{n}}$ the binomial corfficients of three consecutive terms are respectively $220. 495$ and $792$, find the value of n.
AnswerSuppose the three consective terms are $T_{r-1}, T_r$_and $T_{r+1}$.
Coefficients of these terms are ${^\text{n}}\text{C}_{\text{r}-2}, {^\text{n}}\text{C}_{\text{r}-1}$ and ${^\text{n}}\text{C}_{\text{r}},$ respectively.
These corfficients are equal to $220, 495$ and $792$.
$\therefore \frac{{^\text{n}}\text{C}_{\text{r}-2}}{{^\text{n}}\text{C}_{\text{r}-1}}=\frac{220}{495}$
$\Rightarrow \frac{\text{r}-1}{\text{n}-\text{r}+2}=\frac{4}{9}$
$\Rightarrow 4\text{n}+17=13\text{r}\ ...(\text{i})$
Also,
$\therefore \frac{{^\text{n}}\text{C}_{\text{r}}}{{^\text{n}}\text{C}_{\text{r}-1}}=\frac{792}{495}$
$\Rightarrow \frac{\text{n}-\text{r}-1}{\text{r}}=\frac{8}{5}$
$\Rightarrow 5\text{n}-5\text{r}+5=8\text{r}$
$\Rightarrow 5\text{n}+5=13\text{r}$
$\Rightarrow 5\text{n}+5=4\text{n}+17$
$\Rightarrow \text{n}=12$
View full question & answer→Question 185 Marks
Evaluate the following:
$(\sqrt{2}+1)^6+(\sqrt{2}-1)^6$
Answer$(\sqrt{2}+1)^6+(\sqrt{2}-1)^6$
$={^6\text{C}}_0(\sqrt2)^6+{^6\text{C}}_1(\sqrt2)^5+{^6\text{C}}_2(\sqrt2)^4+{^6\text{C}}_3(\sqrt2)^3\\{^6\text{C}}_4(\sqrt2)^2+{^6\text{C}}_5(\sqrt2)+{^6\text{C}}_6+{^6\text{C}}_0(\sqrt2)^6-\\{^6\text{C}}_1(\sqrt2)^5+{^6\text{C}}_2(\sqrt2)^4-{^6\text{C}}_3(\sqrt2)^3+{^6\text{C}}_3(\sqrt2)^3-{^6\text{C}}_4(\sqrt2)^2-{^6\text{C}}_6(\sqrt2)^0$
$=2\big[2^3+15\times2^2+15\times2+1\big]$
$=2\big[8+60+30+1\big]=2(99)=198$
View full question & answer→Question 195 Marks
Find the value of $(1.01)^{10} + (1 - 0.01)^{10}$ correct to $7$ places of decimal.
Answer$(1.01)^{10}+(1-0.01)^{10}=(1-0.01)^{10}+(1-0.01)^{10}$
$=\Big({^{10}\text{C}}_1+{^{10}\text{C}}_2\frac{1}{10^2}+{^{10}\text{C}}_3\frac{1}{10^3}...+{^{10}\text{C}}_{10}\frac{1}{10^{10}}\Big)\\+\Big({^{10}\text{C}}_1-{^{10}\text{C}}_2\frac{1}{10^2}+{^{10}\text{C}}_3\frac{1}{10^3}-{^{10}\text{C}}_4\frac{1}{10^4}+....\Big)$
$=2\Big({^{10}\text{C}}_1-{^{10}\text{C}}_3\frac{1}{10^3}+{^{10}\text{C}}_5\frac{1}{10^5}+{^{10}\text{C}}_7\frac{1}{10^7}+{^{10}\text{C}}_9\frac{1}{10^9}\Big)$
$=2\Big(10+\frac{10!}{3!7!}\frac{1}{1000}+\frac{10!}{5!5!}\frac{1}{(10)^5}+\frac{10!}{7!3!}\times\frac{1}{10^7}+\frac{10!}{9!1!}\frac{1}{10^9}\Big)$
$=2\Big(10+\frac{9\times8}{3\times2\times1000}+\frac{9\times8\times7\times6}{5\times4\times3\times2\times10^5}+\frac{9\times8}{3\times2\times10^7}+\frac{1}{10^8}\Big)$
$=2.00900042$
View full question & answer→Question 205 Marks
The coefficients of 5th, 6th and 7th terms in the expansion of $(1+\text{x})^{\text{n}}$ are in A.P., find n.
AnswerWe have,
$(1+\text{x})^{\text{n}}$
Now,
Coefficient of 5th term $={^\text{n}}\text{C}_{\text{5}-1}={^\text{n}}\text{C}_{\text{4}}$
Coefficient of 5th term $={^\text{n}}\text{C}_{\text{6}-1}={^\text{n}}\text{C}_{\text{5}}$
Coefficient of 5th term $={^\text{n}}\text{C}_{\text{7}-1}={^\text{n}}\text{C}_{\text{6}}$
It is given that these coefficients are in A.P.
$\therefore\ 2\ {^\text{n}}\text{C}_{\text{5}}={^\text{n}}\text{C}_{\text{4}}+{^\text{n}}\text{C}_{\text{6}}$
$\Rightarrow 2\Big[\frac{\text{n}!}{(\text{n}-5)!5!}\Big]=\frac{\text{n}!}{(\text{n}-4)!4!}+\frac{\text{n}!}{(\text{n}-6)!6!}$
$\Rightarrow\ \frac{2}{(\text{n}-5)!5!}=\frac{1}{(\text{n}-4)!4!}+\frac{1}{(\text{n}-6)!6!}$
$\Rightarrow \frac{2}{(\text{n}-5)(\text{n}-6)!5\times4!}=\frac{1}{(\text{4})(\text{n}-5)(\text{n}-6)!4!}+\frac{1}{(\text{n}-6)!6\times5\times4!}$
$\Rightarrow \frac{2}{(\text{n}-5)\times5}=\frac{1}{(\text{n}-4)(\text{n}-5)}+\frac{1}{6\times5}$
$\Rightarrow \frac{2}{5(\text{x}-5)}-\frac{1}{30}=\frac{1}{(\text{n}-4)(\text{n}-5)}$
$\Rightarrow \frac{12-(\text{n}-5)}{30(\text{n}-5)}=\frac{1}{(\text{n}-4)(\text{n}-5)}$
$\Rightarrow \frac{17-\text{n}}{30}=\frac{1}{(\text{n}-4)}$
$\Rightarrow 17\text{n}-68-\text{n}^{2}+4\text{n}=30$
$\Rightarrow 21\text{n}-68-\text{m}^{2}-30=0$
$\Rightarrow 21\text{n}-\text{n}^{2}-98=0$
$\Rightarrow \text{n}^{2}-21\text{n}+98=0$
$\Rightarrow \text{n}^{2}-7\text{n}-14\text{n}+98=0$
$\Rightarrow \text{n}(\text{n}-7)-17(\text{n}-7)=0$
$\Rightarrow (\text{n}-7)(\text{n}-14)=0$
$\Rightarrow \text{n}=7, \text{n}=14$
View full question & answer→Question 215 Marks
Find the term independent of $x$ in the expansion of the following expressions:
$\Big(\sqrt[3]{\text{x}}+\frac{1}{2\sqrt[3]{\text{x}}}\Big)^{18}, \text{x}>2$
AnswerWe have,
$\Big(\sqrt[3]{\text{x}}+\frac{1}{2\sqrt[3]{\text{x}}}\Big)^{18}, \text{x}>0$
Let $(r + 1)^{th}$ term be independent of $x.$
$\text{T}_{\text{r}+1}={^\text{18}}\text{C}_{\text{r}}\big(\sqrt[3]{\text{x}}\big)^{18-\text{r}}+\Big(\frac{1}{2\sqrt[3]{\text{x}}}\Big)^{\text{r}}$
$={^\text{18}}\text{C}_{\text{r}}\Big((\text{x}^\frac{1}{3})\Big)^{18-\text{r}}\times\Big(\frac{1}{2}\Big)^{\text{r}}\times\Big(\frac{1}{\text{x}\frac{1}{3}}\Big)^{\text{r}}$
$={^\text{18}}\text{C}_{\text{r}}(\text{x})^{\frac{18-\text{r}}{3}}\times\Big(\frac{1}{2}\Big)^{\text{r}}$
$={^\text{18}}\text{C}_{\text{r}}(\text{x})^{\frac{18-\text{r}}{3}}\times\Big(\frac{1}{2}\Big)^{\text{r}}$
If it is independent of x, we must have
$\Rightarrow \frac{18-2\text{r}}{3}=0$
$\Rightarrow 18=2\text{r}$
$\Rightarrow \text{r}=9$
Terem independent of $x = T_{9+1} = T_{10}$
Now,
$\text{T}_{10}={^\text{18}}\text{C}_{\text{9}}\big(\sqrt[3]{\text{x}}\big)^{18-9}\Big(\frac{1}{2\sqrt[3]{\text{x}}}\Big)^{9}$
$={^\text{18}}\text{C}_{\text{9}}\big(\sqrt[3]{\text{x}}\big)^{9}\times\frac{1}{2^{9}}\times\Big(\frac{1}{\sqrt[3]{\text{x}}}\Big)^{9}$
$=\frac{{^\text{18}}\text{C}_{\text{9}}}{2^{9}}$
View full question & answer→Question 225 Marks
Find $n$ in the binomial $\Big(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\Big)^{\text{n}},$ if the ratio of $7^{th}$ term from the beginning to the $7^{th}$ term form the end is $\frac{1}{6}.$
AnswerIn the binomial expansion of $\Big(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\Big)^{\text{n}},\big[(\text{n}+1)-7+1\big]^\text{th}\text{i.e.,}\ (\text{n}-5)^{\text{th}}$ term from the beginning is the $7^{th}$ term from the end.
Now,
$\text{T}_{7}={^\text{n}}\text{C}_{\text{6}}\Big(\sqrt[3]{2}\Big)^{\text{n}-6}\Big(\frac{1}{\sqrt[3]{3}}\Big)^{6}={^\text{n}}\text{C}_{\text{6}}\times2^{\frac{\text{n}}{3}-2}\times\frac{1}{3^{2}}$
And,
$\text{T}^{\text{n}-5}={^\text{n}}\text{C}_{\text{n}-6}\Big(\sqrt[3]{2}\Big)^{6}\Big(\frac{1}{\sqrt[3]{3}}\Big)^{\text{n}-6}={^\text{n}}\text{C}_{\text{6}}\times2^{2}\times\frac{1}{3^{\frac{\text{n}}{3}-2}}$
It is given that,
$\frac{\text{T}^{7}}{\text{T}^{\text{n}-5}}=\frac{1}{6}$
$\Rightarrow \frac{{^\text{n}}\text{C}_{\text{6}}\times2^{\frac{\text{n}}{3}-2}\times\frac{1}{3^{2}}}{{^\text{n}}\text{C}_{\text{6}}\times2^{2}\times\frac{1}{3^{\frac{\text{n}}{3}-2}}}=\frac{1}{6}$
$\Rightarrow 2^{\frac{\text{n}}{3}-2-2}\times3^{\frac{\text{n}}{3}-2-2}=\frac{1}{6}$
$\Rightarrow \Big(\frac{1}{6}\Big)^{4-\frac{\text{n}}{3}}=\frac{1}{6}$
$\Rightarrow 4-\frac{\text{n}}{3}=1$
$\Rightarrow \text{n}=9$
Hence, the vaue of $n$ is $9.$
View full question & answer→Question 235 Marks
Find the term independent of x in the expansion of the following expressions:
$\Big(2\text{x}+\frac{1}{3\text{x}^{2}}\Big)^{9}$
AnswerSuppose the (r + 1)th term in the given expression in independent of x.
Now,
$\Big(2\text{x}+\frac{1}{3\text{x}^{2}}\Big)^{9}$
$\text{T}_{\text{r+1}}= {^\text{9}}\text{C}_{\text{r}}\big(2\text{x})^{9-\text{r}}\Big(\frac{1}{3\text{x}^{2}}\Big)^{\text{r}}$
$={^\text{9}}\text{C}_{\text{r}}.\frac{2^{9-\text{r}}}{3^{\text{r}}}\ \text{x}^{\text{9-r}}\Big(\frac{1}{3\text{x}^{2}}\Big)^{\text{r}}$
For ths term to be independent of x, we must have
9 - 36 = 0
⇒ r = 3
Hence, the required term is the 4th term.
Now, we have
${^\text{9}}\text{C}_{\text{3}} \frac{2^{6}}{3^{3}}$
$={^\text{9}}\text{C}_{\text{3}}\times\frac{64}{27}$
View full question & answer→Question 245 Marks
Find the middle terms(s) in the expansion of:
$\Big(\frac{\text{p}}{\text{x}}+\frac{\text{x}}{\text{p}}\Big)^{9}$
AnswerFor the given binomial expansion = 9.
So middle term are $\Big(\frac{9+1}{2}\Big)=5^{\text{th}}$ and $\Big(\frac{9+3}{2}\Big)=6^{\text{th}}$
$\text{T}_{\text{5}}= {^\text{9}}\text{C}_{\text{4}}\Big(\frac{\text{p}}{\text{x}}\Big)^{9-4}\Big(\frac{\text{x}}{\text{p}}\Big)^{4}$
$\text{T}_{\text{5}}= {^\text{9}}\text{C}_{\text{4}}\Big(\frac{\text{p}}{\text{x}}\Big)^{5}\Big(\frac{\text{x}}{\text{p}}\Big)^{4}$
$\text{T}_{\text{5}}= {^\text{9}}\text{C}_{\text{4}}\Big(\frac{\text{p}}{\text{x}}\Big)$
$\text{T}_{\text{5}}= {^\text{9}}\text{C}_{\text{5}}\Big(\frac{\text{p}}{\text{x}}\Big)^{9-5}\Big(\frac{\text{x}}{\text{p}}\Big)^{5}$
$\text{T}_{\text{5}}= {^\text{9}}\text{C}_{\text{5}}\Big(\frac{\text{p}}{\text{x}}\Big)^{4}\Big(\frac{\text{x}}{\text{p}}\Big)^{5}$
$\text{T}_{\text{6}}= {^\text{9}}\text{C}_{\text{5}}\Big(\frac{\text{x}}{\text{p}}\Big)$
$\text{T}_{\text{6}}=\frac{126}{\text{x}}$
The middle terms are $\frac{126}{\text{x}}.$
View full question & answer→Question 255 Marks
If in the coefficients of 2nd, 3rd and 4th terms in the expansion of $(1+\text{x})^{\text{n}}$ are in A.P., then find the value of n.
View full question & answer→Question 265 Marks
Evaluate the following:
$\Big(\text{x}+\sqrt{\text{x}^2-1}\Big)^6+\Big(\text{x}-\sqrt{\text{x}^2-1}\Big)^6$
Answer$\Big(\text{x}+\sqrt{\text{x}^2-1}\Big)^6+\Big(\text{x}-\sqrt{\text{x}^2-1}\Big)^6$
$=2\Big[{^6\text{C}}_0\text{x}^6+{^6\text{C}}_2\text{x}^4\big(\sqrt{\text{x}^2-1}\big)^2+{^6\text{C}}_4\text{x}^2\big(\sqrt{\text{x}^2-1}\big)^4+{^6\text{C}}_6\big(\sqrt{\text{x}^2-1}\big)^6\Big]$
$=2\Big[\text{x}^6+15\text{x}^4\big(\text{x}^2-1\big)+15\text{x}^2\big(\text{x}^2-1\big)^2+\big(\text{x}^2-1\big)^3\Big]$
$=2\Big[\text{x}^6+15\text{x}^6-15\text{x}^4+15\text{x}^6+15\text{x}^2-30\text{x}^4+\text{x}^6-1-3\text{x}^4+3\text{x}^2\Big]$
$=64\text{x}^6-96\text{x}^4+36\text{x}^2-2$
View full question & answer→Question 275 Marks
Using binomial evaluate the following:
$(101)^4$
AnswerWe have,
$(101)^4=(100+1)^4$
$={^4\text{C}}_0\times100^4+{^4\text{C}}_1\times100^3+{^4\text{C}}_2\times100^2+{^4\text{C}}_3\times100+{^4\text{C}}_4$
$=100^4+4\times100^3+6\times100^2+4\times100+1$
$=100000000+4000000+60000+400+1$
$=104060401$
$\therefore(101)^4=104060401$
View full question & answer→Question 285 Marks
Using binomial theorem, prove that $3^{2\text{n}+2}-8\text{n}-9$ is divisible by 64 where $\text{n}\in\text{N}.$
Answer$3^{2\text{n}+2}-8\text{n}-9$
$=3^{2(\text{n}+1)}-8\text{n}-9$
$=9^{\text{n}+1}-8\text{n}-9$
$=(1+8)^{\text{n}+1}-8\text{n}-9$
$\Big({^{\text{n}+1}}\text{C}_0+{^{\text{n}+1}}\text{C}_18^1+{^{\text{n}+1}}\text{C}_28^2+.....+{^{\text{n}+1}}\text{C}_{\text{n}+1}8^{\text{n}+1}\Big)-8\text{n}-9$
$\Big(1+8(\text{n}+1)64^{\text{n}+1}\text{C}_2+64(8)^{\text{n}-1}\Big)-8\text{n}-9$
$=64\Big({^{\text{n}+1}\text{C}}_2+......+8^{\text{n}-1}\Big)$
Thus, $3^{2\text{n}+2}-8\text{n}-9$ is divisible by 64.
View full question & answer→Question 295 Marks
Find the middle terms(s) in the expansion of:
$\Big(2\text{ax}-\frac{\text{b}}{\text{x}^{2}}\Big)^{12}$
Answer$\Big(2\text{ax}-\frac{\text{b}}{\text{x}^{2}}\Big)^{12}$
For the given term is $\Big(\frac{12}{2}+1\Big)=7$
$\text{T}_{\text{7}}= {^\text{12}}\text{C}_{\text{6}}(2\text{ax})^{12-6}\ \Big(\frac{\text{b}}{\text{x}^{2}}\Big)^{6}$
$\text{T}_{\text{7}}= {^\text{12}}\text{C}_{\text{6}}(2\text{ax})^{6}\ \Big(\frac{\text{b}}{\text{x}^{2}}\Big)^{6}$
$\text{T}_{\text{7}}= {^\text{12}}\text{C}_{\text{6}}(2^{6}\text{a}^{6}{\text{x}}^{6})\ \Big(\frac{\text{b}}{\text{x}^{12}}\Big)^{6}$
$\text{T}_{\text{7}}= {^\text{12}}\text{C}_{\text{6}}\Big(\frac{2^{6}\text{a}^{6}\text{b}^{6}}{\text{x}^{6}}\Big)$
View full question & answer→Question 305 Marks
If the term from x in the expansion of $\Big(\sqrt{\text{x}}-\frac{\text{k}}{\text{x}^{2}}\Big)^{10}$ is $405$, find the value of k.
AnswerLet $(r + 1)^{th}$ term, in the expansion of $\Big(\sqrt{\text{x}}-\frac{\text{k}}{\text{x}^{2}}\Big)^{10},$ be free from $x$ and be equal to $T_{r+1}$.
Then,
$\text{T}_{\text{r}+1}={^\text{10}}\text{C}_{\text{r}}(\sqrt{\text{x}})^{10-\text{r}}\Big(\frac{-\text{k}}{\text{x}^{\text{2}}}\Big)^{\text{r}}={^\text{10}}\text{C}_{\text{r}}\text{x}^{5-\frac{5\text{r}}{2}}(-\text{k})^{\text{r}}\ ...(\text{i}) $
If $T_{r+1}$ is independent of $x$, then
$5-\frac{5\text{r}}{2}=0$
$\Rightarrow \text{r}=2$
Putting $r = 2$ in (i), we obtain
$\text{T}_{3}={^\text{10}}\text{C}_{\text{2}}(-\text{k})^{2}=45\text{k}^{2}$
But it is given that the value of the term free $x$ is $405$.
$\therefore 45\text{k}^{2}=405$
$\Rightarrow \text{k}^{2}=9$
$\Rightarrow \text{k}^{2}=\pm3$
Hence, the value of k is $\pm3.$
View full question & answer→Question 315 Marks
Evaluate the following
$\Big(\sqrt{\text{x}+1}+\sqrt{\text{x}-1}\Big)^6+\Big(\sqrt{\text{x}+1}-\sqrt{\text{x}-1}\Big)^6$
Answer$\Big(\sqrt{\text{x}+1}+\sqrt{\text{x}-1}\Big)^6+\Big(\sqrt{\text{x}+1}-\sqrt{\text{x}-1}\Big)^6$
$={^6\text{C}}_0\big(\sqrt{\text{x}+1}\big)^6+{^6\text{C}}_1\big(\sqrt{\text{x}+1}\big)^5\big(\sqrt{\text{x}-1}\big)+{^6\text{C}}_2\big(\sqrt{\text{x}+1}\big)^4\\\big(\sqrt{\text{x}-1}\big)^2-{^6\text{C}}_3\big(\sqrt{\text{x}+1}\big)^3\big(\sqrt{\text{x}-1}\big)^3\\+{^6\text{C}}_4\big(\sqrt{\text{x}+1}\big)^2\big(\sqrt{\text{x}-4}\big)^4+{^6\text{C}_5\big(\sqrt{\text{x}+1}}\big)\big(\sqrt{\text{x}-1}\big)^5+{^6\text{C}}_6\big(\sqrt{\text{x}-1}\big)^6+{^6\text{C}}_0\big(\sqrt{\text{x}+1}\big)^6\\{^6\text{C}}_1\big(\sqrt{\text{x}+1}\big)^5\big(\sqrt{\text{x}-1}\big)+{^6\text{C}}_2\big(\sqrt{\text{x}+1}\big)^4\times\big(\sqrt{\text{x}-1}\big)^2-{^6\text{C}}_3\big(\sqrt{\text{x}+1}\big)^3\\\big(\sqrt{\text{x}-1}\big)63+{^6\text{C}}_4\big(\sqrt{\text{x}+1}\big)^2\big(\sqrt{\text{x}-1}\big)^4-{^6\text{C}}_5\big(\sqrt{\text{x}+1}\big)\big(\sqrt{\text{x}-1}\big){^6\text{C}}_6\big(\sqrt{\text{x}-1}\big)^6$
$=2\big[(\text{x}+1)^3+15(\text{x}+1)^2(\text{x}-1)+15(\text{x}+1)(\text{x}-1)^2+(\text{x}-1)^3\big]$
$=2\Big[\text{x}^3+1+3\text{x}+3\text{x}^2+15\text{x}^3-15\text{x}^2+15\text{x}-15+30\text{x}^2-30\text{x}\\+15\text{x}^3+15\text{x}^2+15\text{x}+15-30\text{x}^2-30\text{x}+\text{x}^3-1-3\text{x}^2+3\text{x}\Big]$
$=64\text{x}^3-48\text{x}$
$=16\text{x}(4\text{x}^2-3)$
View full question & answer→Question 325 Marks
Prove that the coefficient of (r + 1)th term in the expansion of $(1+\text{x})^{\text{n+1}}$ is equal to the sum of the coefficients of rth and (r + 1)th terms in the expansion of $(1+\text{x})^{\text{n}}.$
AnswerCoefficient of (r + 1)th term in the expansion of $(4+\text{x})^{\text{n+1}}={^\text{n+1}}\text{C}_{\text{r}+1-1}={^\text{n+1}}\text{C}_{\text{r}}$
Coefficient of rth term in $(1+\text{x})^{\text{n}}$ + Coefficient of (r + 1)th term in $(1+\text{x})^{\text{n}}$
$={^\text{n}}\text{C}_{\text{r+1}}+{^\text{n}}\text{C}_{\text{r}-1-1}$
$={^\text{n}}\text{C}_{\text{r-1}}+{^\text{n}}\text{C}_{\text{r}}$
$=\frac{\text{n}!}{(\text{n}-(\text{r}-1)!(\text{r}-1)!}+\frac{\text{n}!}{(\text{n}-\text{r})!\text{r}!}$
$=\frac{\text{n}!}{(\text{n}-\text{r}+1)!(\text{r}-1)}+\frac{\text{n}!}{(\text{n}-\text{r})!\text{r}!}$
$=\frac{\text{n}!}{(\text{n}-\text{r}+1)(\text{n}-\text{r})!(\text{r}-1)!}+\frac{\text{n}!}{(\text{n}-\text{r})!\text{r}(\text{r}-1)!}$
$=\frac{\text{n}!}{(\text{n}-\text{r}+1)(\text{n}-\text{r})!(\text{r}-1)!}+\frac{\text{n}!}{(\text{n}-\text{r})!(\text{r}-1)!\text{r}}$
$=\frac{\text{n}!}{(\text{n}-\text{r})!(\text{r}-1)!}\Big[\frac{1}{\text{n}-\text{r}+1}+\frac{1}{\text{r}}\Big]$
$=\frac{\text{n}!}{(\text{n}-\text{r})!(\text{r}-1)!}\Big[\frac{\text{r}+\text{n}-\text{r}+1}{(\text{n}-\text{r}+1)\text{r}}\Big]$
$=\frac{\text{n}!}{(\text{n}-\text{r})!(\text{r}-1)!}\Big[\frac{\text{n+1}}{(\text{n}-\text{r}+1)\text{r}}\Big]$
$=\frac{\text{n}!(\text{n}+1)}{(\text{n}-\text{r})!(\text{n}-\text{r}+1)(\text{r}-1)!\text{r}}$
$=\frac{(\text{n}+1)}{(\text{n}-\text{r}+1)!\text{r}!}$
$=\frac{(\text{n}+1)}{(\text{n}+1-\text{r})!\text{r}!}$
$={^\text{n+1}}\text{C}_{\text{r}}$
${^\text{n+1}}\text{C}_{\text{r}}={^\text{n}}\text{C}_{\text{r}-1}+{^\text{n}}\text{C}_{\text{r}}$
View full question & answer→Question 335 Marks
If p is a real number and if the middle term in the expansion of $\Big(\frac{\text{p}}{2}+2\Big)^{8}$ is 1120, find p.
AnswerIn the binomial expansion of $\Big(\frac{\text{p}}{2}+2\Big)^{8},$ we observe that $\Big(\frac{8}{2}+1\Big)$ i.e., 5th term is the middle term.
It is given that the middle term is 1120.
$\therefore \text{T}_{5}=1120$
$\Rightarrow \therefore\ {^\text{8}}\text{C}_{\text{4}}\Big(\frac{\text{p}}{2}\Big)^{8-4}(2)^{4}=1120$
$\Rightarrow \text{p}^{4}=16$
$\Rightarrow \text{p}=\pm2$
Hence, the real value of p is $\pm2.$
View full question & answer→Question 345 Marks
Find the term independent of x in the expansion of the following expressions:
$\Big(\frac{3\text{x}^{2}}{2}-\frac{1}{3\text{x}}\Big)^{9}$
Answer$\text{T}_{\text{r+1}}= {^\text{9}}\text{C}_{\text{r}}\Big(\frac{3\text{x}^{2}}{2}\Big)^{9-\text{r}}\Big(\frac{-1}{3\text{x}}\Big)^{\text{r}}$
$= {^\text{9}}\text{C}_{\text{r}}\Big(\frac{3}{2}\Big)^{9-\text{r}}\big(\text{x}^{18-2\text{r}}\big)\Big(\frac{-1}{3}\Big)^{\text{r}}\text{x}^{-\text{r}}$
Let $\text{T}_{\text{r+1}}$ be independent of x.
18 - 3r = 0 or r = 6
$\therefore$ Required term,
$\text{T}_{\text{r+1}}=\text{T}_{\text{6+1}}=\text{T}_{\text{7}}= {^\text{9}}\text{C}_{\text{6}}\Big(\frac{3}{2}\Big)^{9-\text{6}}\Big(\frac{-1}{3}\Big)^{\text{6}}\text{x}^{18-3(6)}$
$=84\Big(\frac{27}{8}\Big)\Big(\frac{1}{179}\Big)\text{x}^{0}=\frac{7}{18}$
View full question & answer→Question 355 Marks
Prove that the term independent of x in the expansion of $\Big(\text{x}+\frac{1}{\text{x}}\Big)^{2\text{n}}$ is $\frac{1,3,5.....(2\text{n}-1)}{\text{n}!}.2^{\text{n}}.$
AnswerWe have,
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^{2\text{n}}$
Let $(\text{r}+1)^{\text{2n}}$ term be independent of x.
$\text{T}_{\text{r}+1}={^\text{2n}\text{C}}_\text{r}(\text{x})^{\text{2n}-\text{r}}\Big(\frac{1}{\text{x}}\Big)^{\text{r}}$
$={^\text{2n}\text{C}}_\text{r}(\text{x})^{\text{2n}-\text{r}-\text{r}}$
$={^\text{2n}\text{C}}_\text{r}(\text{x})^{\text{2n}-\text{2r}}$
If it is independent of x, we must have,
$2\text{n}-2\text{r}=0$
$\Rightarrow 2\text{n}=2\text{r}$
$\Rightarrow \text{r}=\text{n}$
Term independent of $\text{x}=\text{T}_{\text{n+1}}$
$={^\text{2n}\text{C}}_\text{n}$
$=\frac{(2\text{n})!}{(2\text{n}-\text{n})!\text{n}!}$
$=\frac{(2\text{n})!}{\text{n}!\text{n}!}$
$=\frac{(2\text{n})(2\text{n}-1)(2\text{n}-2)....5\times4\times3\times2\times1}{\text{n}!\text{n}!}$
$=\frac{1\times3\times5\times.....(2\text{n}-1)(2\times4\times6\times...2\text{n})}{\text{n!}\text{n}!}$
$=\frac{1\times3\times5\times.....(2\text{n}-1)\ \times\ 2^{\text{n}}(1\times2\times3\times...\text{n})}{\text{n!}\text{n}!}$
$=2^{\text{n}}\times\frac{(1\times3\times5\times....(2\text{n}-1))}{\text{n}!}$
The Term independent to $\text{x}=2^{\text{n}}\times\frac{(1\times3\times5\times....(2\text{n}-1))}{\text{n}!}$
Hence proved.
View full question & answer→Question 365 Marks
Find the middle terms(s) in the expansion of:
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^{2\text{n}+1}$
Answer$\Big(\text{x}-\frac{1}{\text{x}}\Big)^{2\text{n}+1}$
2n + 1 is odd hence this expansion will have 2n + 2 = even terms.
Hence, middle terms is $\frac{2\text{n}+1}{2}=\text{n+1},\text{n+2}$
Term formula is,
$\text{T}_{\text{n}}=\text{T}_{\text{r}+1}=(-1)^{r}\ {^\text{n}}\text{C}_{\text{r}}\text{x}^{\text{n-r}}\ \text{y}^{\text{r}}$
$\text{T}_{\text{n+1}}=\text{T}_{\text{n}+1}=(-1)^{\text{n}}\ {^\text{2n+1}}\text{C}_{\text{n}}\text{x}^{\text{2n+1}}\ \Big(\frac{1}{\text{x}}\Big)^{\text{n}}$
$=(-1)^{\text{n}}\ {^\text{2n+1}}\text{C}_{\text{n}}\ \text{x}^{\text{n+1-n}}$
$=(-1)^{\text{n}}\ {^\text{2n+1}}\text{C}_{\text{n}}\text{x}$
$\text{T}_{\text{n+2}}=\text{T}_{\text{n}+1+1}=(-1)^{\text{n}}\ {^\text{2n+1}}\text{C}_{\text{n+1}}\text{x}^{\text{2n+1}-\text{n-1}}\ \Big(\frac{1}{\text{x}}\Big)^{\text{n+1}}$
$=(-1)^{\text{n+1}}\ {^\text{2n+1}}\text{C}_{\text{n+1}}\frac{1}{\text{x}}$
$=(-1)^{\text{n+1}}\ {^\text{2n+1}}\text{C}_{\text{n}}\frac{1}{\text{x}}$
View full question & answer→Question 375 Marks
Evaluate the following:
$(\sqrt3+\sqrt2)^6-(\sqrt3-\sqrt2)^6$
Answer$(\sqrt3+\sqrt2)^6-(\sqrt3-\sqrt2)^6$
$=2\big[{^6\text{C}}_1(\sqrt3)^5(\sqrt2)+{^6\text{C}}_3(\sqrt3)^3(\sqrt2)^3+{^6\text{C}}_5(\sqrt3)(\sqrt2)^5\big]$
$=2\big[6\times\sqrt6\times9+20\times3\sqrt3\times2\sqrt2+6\times\sqrt3\times4\sqrt2\big]$
$=2\big[54\sqrt6+120\sqrt6+24\sqrt6\big]$
$=2\big[198\sqrt6\big]$
$=396\sqrt6$
View full question & answer→Question 385 Marks
Find $(a + b)^4 -(a - b)^4$. Hence, or otherwise evaluate $\Big(\sqrt3+\sqrt2\Big)-\Big(\sqrt3-\sqrt2\Big)$
Answer$(\text{a}+\text{b})^4-(\text{a}-\text{b})^4$
$=\big[{^4\text{C}}_0\text{a}^4\text{b}^0+{^4\text{C}}_1\text{a}^3\text{b}^1+{^4\text{C}}_2\text{a}^2\text{b}^2{\text{C}}_3\text{a}^1\text{b}^3+{^4\text{C}}_4\text{a}^0\text{b}^4\big]\\-\big[{^4\text{C}}_0\text{a}^4\text{b}^0+{^4\text{C}}_1\text{a}^3\text{b}^1+{^4\text{C}}_2\text{a}^2\text{b}^2-{^4\text{C}}_3\text{a}^1\text{b}^3+{^4\text{C}}_4\text{a}^0\text{b}^4\big]$
$=\big[{^4\text{C}}_0\text{a}^4(-\text{b})^0+{^4\text{C}}_1\text{a}^3(-\text{b})^1+{^4\text{C}}_2\text{a}^2(-\text{b})+{^4\text{C}}_3\text{a}^1(-\text{b})^3+{^4\text{C}}_4\text{a}^0(-\text{b})^4\big]\\-\big[{^4\text{C}}_0\text{a}^4(-\text{b})^0+{^4\text{C}}_1\text{a}^3(-\text{b})^1+{^4\text{C}}_2\text{a}^2(-\text{b})^2+{^4\text{C}}_3\text{a}^1(-\text{b})^3+{^4\text{C}}_4\text{a}^0(-\text{b})^4\big]$
$=\big[{^4\text{C}}_0\text{a}^4+{^4\text{C}}_1\text{a}^3\text{b}+{^4\text{C}}_2\text{a}^2\text{b}^2+{^4\text{C}}_3\text{ab}^3+{^4\text{C}}_4\text{ab}^4\big]\\-\big[{^4\text{C}}_0\text{a}^4-{^4\text{C}}_1\text{a}^3\text{b}+{^4\text{C}}_2\text{a}^2\text{b}^2-{^4\text{C}}_3\text{ab}^3+{^4\text{C}}_4\text{b}^4$
$={^4\text{C}}_0\text{a}^4+{^4\text{C}}_1\text{a}^3\text{b}+{^4\text{C}}_2\text{a}^2\text{b}^2+{^4\text{C}}_3\text{ab}^3+{^4\text{C}}_4\text{ab}^4\\-{^4\text{C}}_0\text{a}^4+{^4\text{C}}_1\text{a}^3\text{b}-{^4\text{C}}_2\text{a}^2\text{b}^2+{^4\text{C}}_3\text{ab}^3+{^4\text{C}}_4\text{b}^4\big]$
$=2\big[{^4\text{C}}_1\text{a}^3\text{b}+{^4\text{C}}_3\text{ab}^3\big]$
$=2\big[4\text{a}^3\text{b}+4\text{ab}^3\big]$
$=8\big[\text{a}^3\text{b}+\text{ab}^3\big]$
$\therefore(\text{a+b})^4-(\text{a-b})^4=8(\text{a}^3\text{b}+\text{ab}^3)$
Putting $\text{a}=\sqrt3$ and $\text{b}=\sqrt2$ in equation (i) we get
$(\sqrt3+\sqrt2)^4-(\sqrt3-\sqrt2)^4=8\Big[(\sqrt3)^3\times\sqrt2+(\sqrt3)\times(\sqrt2)^2\Big]$
$=8\Big[3\sqrt6+2\sqrt6\Big]$
$=8\times5\sqrt6$
$=40\sqrt6$
$\therefore(\sqrt{3}+\sqrt2)^4-(\sqrt{3}-\sqrt2)^440\sqrt6.$
View full question & answer→Question 395 Marks
Find the term independent of $x$ in the expansion of the following expressions:
$\Big(2\text{x}^{2}-\frac{3}{\text{x}^{3}}\Big)^{25}$
Answer$\text{T}_{\text{r+1}}=(-1)^{\text{r}} \ {^\text{n}}\text{C}_{\text{r}}\big(2\text{x}^{2})^{25-\text{r}}\Big(\frac{3}{\text{x}^{2}}\Big)^{\text{r}}$
$=(-1)^{\text{r}} \ {^\text{n}}\text{C}_{\text{r}}\big(2)^{25-\text{r}}\ 3^{\text{r}}\ \text{x}^{50-2\text{r}-3\text{r}}$
Term independent of $x = x^0$
$\Rightarrow \text{x}^{50-50}=\text{x}^{0}$
$\Rightarrow 50-5\text{r}=0$
$\Rightarrow \text{r}=10$
$\text{T}_{11}=(-1)^{\text{10}} \ {^\text{25}}\text{C}_{\text{10}}\big(2)^{15}\times 3^{\text{10}}$
$={^\text{25}}\text{C}_{\text{10}}\big(2)^{15}\times 3^{\text{10}}$
View full question & answer→Question 405 Marks
Find $a, b$ and $n$ in the expansion of $(\text{a}+\text{b})^{\text{n}},$ if the first three terms in the expansion are $729, 7290$ and $30375$ respectively.
AnswerWe have,
$T_1 = 729, T_2 = 7290$ and $T_3 = 30375$
According to the quation,
${^\text{n}}\text{C}_{\text{0}}\text{a}^{\text{n}}\text{b}^{0}=729$
$\Rightarrow \text{a}^{\text{n}}=729$
$\Rightarrow \text{a}^{\text{n}}=3^{6}$
${^\text{n}}\text{C}_{\text{1}}\text{a}^{\text{n}-1}\text{b}^{1}=7290$
${^\text{n}}\text{C}_{\text{2}}\text{x}^{\text{n}-2}\text{a}^{2}=30375$
Also,
$\Rightarrow \frac{{^\text{n}}\text{C}_{\text{2}}\text{x}^{\text{n}-2}\text{a}^{2}}{{^\text{n}}\text{C}_{\text{1}}\text{x}^{\text{n}-1}\text{a}^{1}}=\frac{30375}{7290}$
$\Rightarrow \frac{\text{n}-1}{2}\times\frac{\text{b}}{\text{a}}=\frac{25}{6}\ ...(\text{i})$
$\Rightarrow \frac{\big(\text{n}-1\big)\text{b}}{\text{a}}=\frac{25}{3}$
And,
$\frac{{^\text{n}}\text{C}_{\text{1}}\text{a}^{\text{n}-1}\text{b}^{1}}{{^\text{n}}\text{C}_{\text{0}}\text{a}^{\text{n}}\text{b}^{0}}=\frac{7290}{729}$
$\Rightarrow \frac{\text{n}\text{b}}{\text{a}}=\frac{10}{1}\ ...(\text{ii})$
On dividing (ii) by (i), we get
$\frac{\frac{\text{nd}}{\text{a}}}{\frac{(\text{n}-1)\text{b}}{\text{a}}}=\frac{10\times3}{25}$
$\Rightarrow \frac{\text{n}}{\text{n}-1}=\frac{6}{5}$
$\Rightarrow \text{n}=6$
Since, $\text{a}^{6}=3^{6}$
Hence, $\text{a}=3$
Now,
$\frac{\text{nb}}{\text{a}}=10$
$\Rightarrow \text{b}=5$
View full question & answer→Question 415 Marks
Evaluate the following:
$(3+\sqrt2)^5-(3-\sqrt2)^5$
Answer$(3+\sqrt2)^5-(3-\sqrt2)^5$
$=2\big[{^5\text{C}}_1(3)^4(\sqrt2)^1+{^5\text{C}}_3(3)^2(\sqrt2)^3+{^5\text{C}}_5(\sqrt3)^5\big]$
$=2\big[5\times81\times\sqrt2+10\times9\times2\sqrt2+4\sqrt2\big]$
$=2\big[405\sqrt2+180\sqrt2+4\sqrt2\big]$
$=2\big[589\sqrt2\big]$
$=1178\sqrt2$
View full question & answer→Question 425 Marks
Find the middle terms(s) in the expansion of:
$\Big(\frac{\text{x}}{\text{a}}-\frac{\text{a}}{\text{x}}\Big)^{10}$
AnswerFor the given binomial expansion = 10.
So middle term are $\Big(\frac{10}{2}+1\Big)=6^{\text{th}}$
$\text{T}_{\text{5}}= {^\text{10}}\text{C}_{\text{5}}\Big(\frac{\text{p}}{\text{x}}\Big)^{10-5}\Big(-\frac{\text{a}}{\text{x}}\Big)^{5}$
$\text{T}_{\text{6}}= -{^\text{10}}\text{C}_{\text{5}}\Big(\frac{\text{x}}{\text{a}}\Big)^{5}\Big(\frac{\text{a}}{\text{x}}\Big)^{5}$
$\text{T}_{\text{5}}= {^\text{10}}\text{C}_{\text{5}}=-252$
The middle terms are -252.
View full question & answer→Question 435 Marks
The coefficients of 2nd, 3rd and 4th terms in the expansion of $(1+\text{x})^{2\text{n}}$ are in A.P., then show that $2\text{n}^{2}-9\text{n}+7=0.$
AnswerWe have,
$(1+\text{x})^{\text{2n}}$
Now,
Coefficient of 2nd term $={^\text{2n}}\text{C}_{\text{2}-1}={^\text{2n}}\text{C}_{\text{1}}$
Coefficient of 3rd term $={^\text{2n}}\text{C}_{\text{3}-1}={^\text{2n}}\text{C}_{\text{2}}$
Coefficient of 4th term $={^\text{2n}}\text{C}_{\text{4}-1}={^\text{2n}}\text{C}_{\text{3}}$
It is given that these coefficients are in A.P.
$\therefore\ 2\ {^\text{2n}}\text{C}_{\text{2}}={^\text{2n}}\text{C}_{\text{1}}+{^\text{2n}}\text{C}_{\text{3}}$
$\Rightarrow 2=\frac{{^\text{2n}}\text{C}_{\text{1}}}{{^\text{2n}}\text{C}_{\text{2}}}+\frac{{^\text{2n}}\text{C}_{\text{3}}}{{^\text{2n}}\text{C}_{\text{2}}}$
$\Rightarrow\ 2=\frac{2}{2\text{n}-2+1}+\frac{2\text{n}-3+1}{3}$
$\Rightarrow 2=\frac{6+(2\text{n}+1)(2\text{n}-2)}{3(2\text{n}-1)}$
$\Rightarrow 6(2\text{n}-1)=6+4\text{n}^{2}-4\text{n}-2\text{n}+2$
$\Rightarrow 12\text{n}-6=8+4\text{n}^{2}-6\text{n}$
$\Rightarrow 4\text{n}^{2}-6\text{n}-12\text{n}+8+6=0$
$\Rightarrow 4\text{n}^{2}-18\text{n}+14=0$
$\Rightarrow 2(2\text{n}^{2}-9\text{n}+7)=0$
$\Rightarrow 2\text{n}^{2}-9\text{n}+7=0$
Hence proved.
View full question & answer→Question 445 Marks
If the coefficients of $(2r + 4)$th and $(r - 2)$th teram in the expansion of $(1 + x)^{18}$ are equal, find r.
AnswerWe know that the coefficient of rth term in the expansion of $(1+\text{x}){^\text{n}}$ is ${^\text{n}}\text{C}_{\text{r}-1}$
Coefficient of $(2r + 4)$th term of the expansion $(1+\text{x})^{18}={^\text{18}}\text{C}_{\text{2r}+4-1}={^\text{18}}\text{C}_{\text{2r}+3}$
Coefficient of $(r - 2)$th term of the expansion $(1+\text{x})^{18}={^\text{18}}\text{C}_{\text{r}-2-1}={^\text{18}}\text{C}_{\text{r}-3}$
It is given that these coefficients are equal.
$\Rightarrow {^\text{18}}\text{C}_{\text{2r}+3}={^\text{18}}\text{C}_{\text{r}-3}$
$\Rightarrow 2\text{r}+3=\text{r}-3$ or $2\text{r}+3+\text{r}-3=18$
$\Rightarrow \text{r}=-6$ or $\text{r}=6$
$\Rightarrow\text{r}=6$
View full question & answer→Question 455 Marks
Evaluate the following:
$(0.99)^5+(1.01)^5$
Answer$(0.99)^5+(1.01)^5$
$=(1-.01)^5+(1+.01)^5$
$=2\big[{^5\text{C}}_1+{^5\text{C}}_3(.01)^2+{^5\text{C}}_5(.01)^5\big]$
$=2\Big[5+10\times\frac{1}{10^4}+\frac{1}{10^{10}}\Big]$
$=2\Big[5+\frac{1}{1000}+\frac{1}{10^{10}}\Big]$
$=2.0020001$
View full question & answer→Question 465 Marks
Find $(\text{x}+1)^6+(\text{x}-1)^6.$ Hence or otherwise evaluate $(\sqrt2+1)^6+(\sqrt2-1).$
AnswerWe have,
$(\text{x}+1)^6+(\text{x}-1)^6$
$=\Big[{^6\text{C}}_0\text{x}^6+{^6\text{C}}_1\text{x}^5+{^6\text{C}}_2\text{x}^4+{^6\text{C}}_3\text{x}^3+{^6\text{C}}_4\text{x}^2+{^6\text{C}}_5\text{x}^1+{^6\text{C}}_6\text{X}^0\big]\\+\big[{^6\text{C}}_0\text{x}^6(-1)^0+{^6\text{C}}_1\text{x}^5(-1)^1+{^6\text{C}}_2\text{x}^4(-1)^2+{^6\text{C}}_3\text{x}^3(-1)^3\\+{^6\text{C}}_4\text{x}^2(-1)^4+{^6\text{C}}_5\text{x}^1(-1)^5+{^6\text{C}}_6\text{x}^0(-1)^6\big]\\{^6\text{C}}_0\text{x}^6-{^6\text{C}}_1\text{x}^5+{^6\text{C}}_2\text{x}^4-{^6\text{C}}_3\text{x}^3+{^6\text{C}}_4\text{x}^2-{^6\text{C}}_5\text{x}+{^\text{C}}_6\Big]$
$=\Big[{^6\text{C}}_0\text{x}^6+{^6\text{C}}_1\text{x}^5+{^6\text{C}}_2\text{x}^4+{^6\text{C}}_3\text{x}^3+{^6\text{C}}_4\text{x}^2+{^6\text{C}}_5\text{x}+{^6\text{C}}_6\\{^6\text{C}}_0\text{x}^6-{^6\text{C}}_1\text{x}^5+{^6\text{C}}_2\text{x}^4-{^6\text{C}}_3\text{x}^3+{^6\text{C}}_4\text{x}^2-{^6\text{C}}_5\text{x}+{^6\text{C}}_6\Big]$
$=2\big[{^6\text{C}}_0\text{x}^6+{^6\text{C}}_2\text{x}^4+{^6\text{C}}_4\text{x}^2+{^6\text{C}}_6\big]$
$=2\big[\text{x}^6+15\text{x}^4+15\text{x}^2+1\big]$
$\therefore(\text{x}+1)^6+(\text{x}-1)^6=2\big[\text{x}^6+15\text{x}^4+15\text{x}^2+1\big]...(\text{i})$
Putting $\text{x}=\sqrt2$ in equation (i), we get
$(\text{x}+1)^6+(\text{x}-1)^6=2\big[(\sqrt2)^6+15(\sqrt2)^4+15(\sqrt2)^2+1\big]$
$=2\big[8+60+30+1\big]$
$=2\big[99\big]$
$=198$
$\therefore(\text{x}+1)^6+(\text{x}-1)^6=198$
View full question & answer→Question 475 Marks
Using binomial theorem, indicate which is larger $(1.1)^{10000}$ or $1000?$
AnswerWe have,
$(1.1)^{10000}=(1+0.1)^{10000}$
$={^{10000}\text{C}}_0+{^{10000}\text{C}}_1(0.1)+{^{10000}\text{C}}_2(0.1)^2+...+{^{10000}\text{C}}_{10000}(0.1)^{10000}$
$= 1 + 10000 \times (0.1) +$ other positive terms
$= 1 + 1000 +$ other positive terms
$= 1001 +$ other positive terms $> 1000$
$\therefore (1.1)^{10000}> 1000$
View full question & answer→Question 485 Marks
Using binomial theorem determine which number is smaller $(1.2)^{4000}$ or $800?$
Answer$(1.2)^{4000}=(1+0.2)^{4000}$
$={^{4000}\text{C}}_0(0.2)^0(1)^{4000}+{^{4000}\text{C}}_1(0.2)^1{1}^{3999}+....\\+{^{4000}\text{C}}_400(0.2)^0(1)^{4000}(0.2)^{4000}1^0$
$=1 + 4000 \times 0.2 \times 1 +.......+(0.2)^{4000}$
$= 1 + 800 +.......+(0.2)^{4000}$
Here, we dearty observe $(1, 2)^{4000} $ is less than $(801)$ thus, $(1, 2)^{4000} < 800.$
View full question & answer→Question 495 Marks
Using binomial evaluate the following:
$(96)^3$
AnswerWe have,
$(96)^3=(100-4)^3$
$={^3\text{C}}_0\times100^3+{^3\text{C}}_1\times100^2\times(-4)+{^3\text{C}}_2\times100\times(-4)^2+{^3\text{C}}_3\times(-4)^3$
$=100^3-3\times100^2\times4+3\times100\times4^2-4^3$
$=1000000-120000+4800-64$
$1004800-120064$
$=884736$
$\therefore(96)^3=884736$
View full question & answer→Question 505 Marks
Using binomial theorem, prove that $2^{3\text{n}}-7\text{n}-1$ is divisible by 49 where $\text{n}\in\text{N}.$
Answer$2^{3\text{n}}-7\text{n}-1$
$=2^{3(\text{n})}-7(\text{n})-1$
$=8^\text{n}-7\text{n}-1$
$=(1+7)^\text{n}-7\text{n}-1$
$==({^\text{n}\text{C}}_0+{^\text{n}\text{C}}_1(7)^1+{^\text{n}\text{C}}_2(7)^2+....{^\text{n}\text{C}}_\text{n}(7)^\text{n})-7\text{n}-1$
$=(1+7\text{n}+49^\text{n}\text{C}_2+......49(7)^{\text{n}-2})-7\text{n}-1$
$=49({^\text{n}\text{C}}_2+......+7^{\text{n}-2})$
$\therefore2^{3\text{n}}-7\text{n}-1$ is divisible by 49
Hence, proved
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