1. Find $a$,$b$ and $n$ in the expansion of $\displaystyle \small (a+b)^{n}$ if the first three terms of the expansion are 729, 7290 and 30375, respectively.
We know that,
$\displaystyle \small (a+b)^{n}=^{n}C_{0}a^{n}+^{n}C_{1}a^{n-1}b$$\displaystyle \small +^{n}C_{2}a^{n-2}b^{2}+...$
Given the values of first three terms of the exapnsion.
∴ $\displaystyle \small ^{n}C_{0}a^{n}=729$
$\displaystyle \small ^{n}C_{1}a^{n-1}b=7290$
$\displaystyle \small ^{n}C_{2}a^{n-2}b^{2}=30375$
$\displaystyle \small ^{n}C_{0}a^{n}=729$
⇒ $\displaystyle \small a^{n}=729$ ...(i)
$\displaystyle \small ^{n}C_{1}a^{n-1}b=7290$
⇒ $\displaystyle \small \frac{n!}{1!(n-1)!}a^{n-1}b=7290$
$\displaystyle \small \frac{n(n-1)!}{(n-1)!}a^{n}a^{-1}b=7290$
$\displaystyle \small n \frac{a^{n}}{a}b=7290$
$\displaystyle \small n \frac{729}{a}b=7290$ [from (i)]
$\displaystyle \small n \frac{b}{a}=\frac{7290}{729}$
$\displaystyle \small n \frac{b}{a}=10$ ...(ii)
$\displaystyle \small ^{n}C_{2}a^{n-2}b^{2}=30375$
⇒ $\displaystyle \small \frac{n!}{2!(n-2)!}a^{n}a^{-2}b^{2}=30375$
$\displaystyle \small \frac{n(n-1)(n-2)!}{2(n-2)!}a^{n}a^{-2}b^{2}=30375$
$\displaystyle \small \frac{n(n-1)}{2}\frac{a^{n}}{a^{2}}b^{2}=30375$
$\displaystyle \small \frac{n(n-1)}{2}\frac{729}{a^{2}}b^{2}=30375$
$\displaystyle \small n(n-1)\frac{b^{2}}{a^{2}}=\frac{30375*2}{729}=\frac{250}{3}$ ...(iii)
Divide (iii) by square of (ii),
⇒ $\displaystyle \small \frac{n(n-1)\frac{b^{2}}{a^{2}}}{(n \frac{b}{a})^{2}}=\frac{\frac{250}{3}}{10^2}$
$\displaystyle \small \frac{n(n-1)\frac{b^{2}}{a^{2}}}{n^{2} \frac{b^{2}}{a^{2}}}=\frac{\frac{250}{3}}{100}$
$\displaystyle \small \frac{(n-1)}{n}=\frac{250}{300}$
$\displaystyle \small \frac{n-1}{n}=\frac{5}{6}$
$\displaystyle \small 6n-6=5n$
$\displaystyle \small n=6$
(i) ⇒ $\displaystyle \small a^{6}=729$
$\displaystyle \small 3^{6}=729$
$\displaystyle \small a=3$
(ii) ⇒ $\displaystyle \small n \frac{b}{a}=10$
$\displaystyle \small 6 \frac{b}{3}=10$
$\displaystyle \small b=5$
2. Find $a$ if the coefficients of $x^{2}$ and $x^{3}$ in the expansion of $\displaystyle \small (3+ax)^{9}$ are equal.
general expansion of $\displaystyle \small (3+ax)^{9}$ is,
$\displaystyle \small T_{r+1}=^{9}C_{r}(3)^{9-r}(ax)^{r}$
Put $r=2$ in (i),
⇒ $\displaystyle \small T_{2+1}=^{9}C_{2}(3)^{9-2}(ax)^{2}$
$\displaystyle \small T_{3}=^{9}C_{2}(3)^{7}a^{2}x^{2}$
Put $r=3$ in (i),
⇒ $\displaystyle \small T_{3+1}=^{9}C_{3}(3)^{9-3}(ax)^{3}$
$\displaystyle \small T_{4}=^{9}C_{3}(3)^{6}a^{3}x^{3}$
Given that coefficients of $x^{2}$ and $x^{3}$ are equal.
⇒ $\displaystyle \small ^{9}C_{2}(3)^{7}a^{2}=^{9}C_{3}(3)^{6}a^{3}$
$\displaystyle \small (36)(3)^{7}a^{2}=(84)(3)^{6}a^{3}$
$\displaystyle \small (36)(3)=(84)a$
$\displaystyle \small a=\frac{9}{7}$
3. Find the coefficient of $x^{5}$ in the product $\displaystyle \small (1+2x)^{6}(1-x)^{7}$ using binomial theorem.
$\displaystyle \small (1+2x)^{6}(1-x)^{7}$ =$\displaystyle \small [^{6}C_{0}+^{6}C_{1}(2x)+^{6}C_{2}(2x)^{2}+^{6}C_{3}(2x)^{3}$$\displaystyle \small +^{6}C_{4}(2x)^{4}+^{6}C_{5}(2x)^{5}+^{6}C_{6}(2x)^{6}]$ $\displaystyle \small [^{7}C_{0}-^{7}C_{1}x+^{7}C_{2}x^{2}-^{7}C_{3}x^{3}$$\displaystyle \small +^{7}C_{4}x^{4}-^{7}C_{5}x^{5}+^{7}C_{6}x^{6}-^{7}C_{7}x^{7}]$
= $\displaystyle \small [1+6(2x)+15(2x)^{2}+20(2x)^{3}+15(2x)^{4}$$\displaystyle \small +6(2x)^{5}+(2x)^{6}]$ $\displaystyle \small [1-7x+21x^{2}-35x^{3}+35x^{4}-21x^{5}+7x^{6}$$\displaystyle \small -x^{7}]$
= $\displaystyle \small [1+12x+60x^{2}+160x^{3}+240x^{4}+192x^{5}$$\displaystyle \small +64x^{6}]$ $\displaystyle \small [1-7x+21x^{2}-35x^{3}+35x^{4}-21x^{5}+7x^{6}$$\displaystyle \small -x^{7}]$
coefficient of $x^{5}$ in the product is,
= $\displaystyle \small [-1*21+12*35-60*35+160*21$$\displaystyle \small -240*7+192*1]$
= $\displaystyle \small -21+420-2100+3360-1680+192$$\displaystyle \small =171$
4. If $a$ and $b$ are distinct integers, prove that $a-b$ is a factor of $a^{n}-b^{n}$, whenever n is a positive integer. [Hint: write $a^{n}=(a-b+b)^{n}$ and expand]
Let $\displaystyle \small a^{n}=((a-b)+b)^{n}$ = $\displaystyle \small ^{n}C_{0}(a-b)^{n}+^{n}C_{1}(a-b)^{n-1}b$$\displaystyle \small +^{n}C_{2}(a-b)^{n-2}b^{2}+...+^{n}C_{n-1}(a-b)b^{n-1}$$\displaystyle \small +^{n}C_{n}b^{n}$
$\displaystyle \small a^{n}=(a-b)^{n}+^{n}C_{1}(a-b)^{n-1}b$$\displaystyle \small +^{n}C_{2}(a-b)^{n-2}b^{2}+...+^{n}C_{n-1}(a-b)b^{n-1}$$\displaystyle \small +b^{n}$
$\displaystyle \small a^{n}-b^{n}=(a-b)^{n}+^{n}C_{1}(a-b)^{n-1}b$$\displaystyle \small +^{n}C_{2}(a-b)^{n-2}b^{2}+...+^{n}C_{n-1}(a-b)b^{n-1}$
$\displaystyle \small a^{n}-b^{n}=(a-b)[(a-b)^{n-1}$$\displaystyle \small +^{n}C_{1}(a-b)^{n-2}b+^{n}C_{2}(a-b)^{n-2}b^{3}+...$$\displaystyle \small +^{n}C_{n-1}b^{n-1}]$
which shows $a-b$ is a factor of $a^{n}-b^{n}$
5. Evaluate $\displaystyle \small (\sqrt{3}+\sqrt{2})^{6}-(\sqrt{3}-\sqrt{2})^{6}$
$\displaystyle \small (\sqrt{3}+\sqrt{2})^{6}-(\sqrt{3}-\sqrt{2})^{6}$ = $\displaystyle \small [^{6}C_{0}(\sqrt{3})^{6}+^{6}C_{1}(\sqrt{3})^{5}\sqrt{2}$$\displaystyle \small +^{6}C_{2}(\sqrt{3})^{4}(\sqrt{2})^{2}$ $\displaystyle \small +^{6}C_{3}(\sqrt{3})^{3}(\sqrt{2})^{3}+^{6}C_{4}(\sqrt{3})^{2}(\sqrt{2})^{4}$$\displaystyle \small +^{6}C_{5}\sqrt{3}(\sqrt{2})^{5}+^{6}C_{6}(\sqrt{2})^{6}]$ $\displaystyle \small -[^{6}C_{0}(\sqrt{3})^{6}-^{6}C_{1}(\sqrt{3})^{5}\sqrt{2}+^{6}C_{2}(\sqrt{3})^{4}(\sqrt{2})^{2}$$\displaystyle \small -^{6}C_{3}(\sqrt{3})^{3}(\sqrt{2})^{3}$ $\displaystyle \small +^{6}C_{4}(\sqrt{3})^{2}(\sqrt{2})^{4}$$\displaystyle \small -^{6}C_{5}\sqrt{3}(\sqrt{2})^{5}+^{6}C_{6}(\sqrt{2})^{6}]$
= $\displaystyle \small 2(^{6}C_{1}(\sqrt{3})^{5}\sqrt{2})+2(^{6}C_{3}(\sqrt{3})^{3}(\sqrt{2})^{3})$$\displaystyle \small +2(^{6}C_{5}\sqrt{3}(\sqrt{2})^{5})$
= $\displaystyle \small 2(6)(\sqrt{3})^{5}\sqrt{2}+2(20)(\sqrt{3})^{3}(\sqrt{2})^{3}$$\displaystyle \small +2(6)\sqrt{3}(\sqrt{2})^{5}$
= $\displaystyle \small 2(6)(9)\sqrt{3}\sqrt{2}+2(20)(3)\sqrt{3}(2)\sqrt{2}$$\displaystyle \small +2(6)\sqrt{3}(4)\sqrt{2}$
= $\displaystyle \small 108\sqrt{6}+240\sqrt{6}+48\sqrt{6}=396\sqrt{6}$
6. Find the value of $\displaystyle \small (a^{2}+\sqrt{a^{2}-1})^{4}+(a^{2}-\sqrt{a^{2}-1})^{4}$
$\displaystyle \small (a^{2}+\sqrt{a^{2}-1})^{4}+(a^{2}-\sqrt{a^{2}-1})^{4}$= $\displaystyle \small [^{4}C_{0}(a^{2})^{4}+^{4}C_{1}(a^{2})^{3}(\sqrt{a^{2}-1})$$\displaystyle \small +^{4}C_{2}(a^{2})^{2}(\sqrt{a^{2}-1})^{2}$ $\displaystyle \small +^{4}C_{3}a^{2}(\sqrt{a^{2}-1})^{3}+^{4}C_{4}(\sqrt{a^{2}-1})^{4}]$$\displaystyle \small +[^{4}C_{0}(a^{2})^{4}-^{4}C_{1}(a^{2})^{3}(\sqrt{a^{2}-1})$ $\displaystyle \small +^{4}C_{2}(a^{2})^{2}(\sqrt{a^{2}-1})^{2}-^{4}C_{3}a^{2}(\sqrt{a^{2}-1})^{3}$$\displaystyle \small +^{4}C_{4}(\sqrt{a^{2}-1})^{4}]$
= $\displaystyle \small 2(^{4}C_{0}(a^{2})^{4})+2(^{4}C_{2}(a^{2})^{2}(\sqrt{a^{2}-1})^{2})$$\displaystyle \small +2(^{4}C_{4}(\sqrt{a^{2}-1})^{4})$
= $\displaystyle \small 2(a^{2})^{4}+2(6)(a^{2})^{2}(\sqrt{a^{2}-1})^{2}$$\displaystyle \small +2(\sqrt{a^{2}-1})^{4}$
= $\displaystyle \small 2a^{8}+12a^{4}(a^{2}-1)+2(a^{2}-1)^{2}$
= $\displaystyle \small 2[a^{8}+6a^{4}(a^{2}-1)+(a^{2}-1)^{2}]$
= $\displaystyle \small 2[a^{8}+6a^{6}-6a^{4}+a^{4}+1-2a^{2}]$
= $\displaystyle \small 2[a^{8}+6a^{6}-5a^{4}-2a^{2}+1]$
7. Find an approximation of $(0.99)^{5}$ using the first three terms of its expansion.
$\displaystyle \small (0.99)^{5}=(1-0.01)^{5}$
= $\displaystyle \small ^{5}C_{0}-^{5}C_{1}(0.01)+^{5}C_{2}(0.01)^{2}+...$
= $\displaystyle \small 1-5(0.01)+10(0.01)^{2}+...$
= $\displaystyle \small 1-0.05+0.001+...=0.951$
8. Find n if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of $\displaystyle \small \left ( \sqrt[4]{2}+\frac{1}{\sqrt[4]{3}} \right )^{n}$ is $\displaystyle \small \sqrt{6}:1$ .
We know that, $\displaystyle \small T_{r+1}=^{n}C_{r}a^{n-r}b^{r}$
fifth term from the beginning is when $r=4$
fifth term from the end is when $r=n-4$
∴ $\displaystyle \small T_{4+1}=^{n}C_{4} \left (\sqrt[4]{2} \right )^{n-4} \left ( \frac{1}{\sqrt[4]{3}} \right )^{4}$
$\displaystyle \small T_{n-4+1}=^{n}C_{n-4} \left (\sqrt[4]{2} \right )^{n-(n-4)} \left ( \frac{1}{\sqrt[4]{3}} \right )^{n-4}$
Given that, $\displaystyle \small T_{4+1}:T_{n-4+1}=\sqrt{6}:1$
$\displaystyle \small \frac{^{n}C_{4} \left (\sqrt[4]{2} \right )^{n-4} \left ( \frac{1}{\sqrt[4]{3}} \right )^{4}}{^{n}C_{n-4} \left (\sqrt[4]{2} \right )^{4} \left ( \frac{1}{\sqrt[4]{3}} \right )^{n-4}}=\frac{\sqrt{6}}{1}$
$\displaystyle \small \frac{ \left (\sqrt[4]{2} \right )^{n-4} \left ( \frac{1}{\sqrt[4]{3}} \right )^{4}}{ \left (\sqrt[4]{2} \right )^{4} \left ( \frac{1}{\sqrt[4]{3}} \right )^{n-4}}=\sqrt{6}$ [$^{n}C_{4}=^{n}C_{n-4}$]
$\displaystyle \small \frac{ \left (\sqrt[4]{2} \right )^{n-4} \left ( \sqrt[4]{3} \right )^{-4}}{ \left (\sqrt[4]{2} \right )^{4} \left (\sqrt[4]{3} \right )^{-n+4}}=\sqrt{6}$
$\displaystyle \small \left (\sqrt[4]{2} \right )^{n-4-4} \left ( \sqrt[4]{3} \right )^{-4+n-4}=\sqrt{6}$
$\displaystyle \small \left (\sqrt[4]{2} \right )^{n-8} \left ( \sqrt[4]{3} \right )^{n-8}=\sqrt{6}$
$\displaystyle \small (2)^\frac{n-8}{4}(3)^\frac{n-8}{4}=\sqrt{6}$
$\displaystyle \small (6)^\frac{n-8}{4}=(6)^\frac{1}{2}$
$\displaystyle \small \frac{n-8}{4}=\frac{1}{2}$
$\displaystyle \small n-8=2$
$\displaystyle \small n=10$
9. Expand using Binomial theorem $\displaystyle \small \left ( 1+\frac{x}{2}-\frac{2}{x} \right )^{4}$
$\displaystyle \small \left ( 1+\frac{x}{2}-\frac{2}{x} \right )^{4}=\left ( 1+\left (\frac{x}{2}-\frac{2}{x} \right ) \right )^{4}$
= $\displaystyle \small ^{4}C_{0}+^{4}C_{1}\left (\frac{x}{2}-\frac{2}{x} \right )+^{4}C_{2}\left (\frac{x}{2}-\frac{2}{x} \right )^{2}$$\displaystyle \small +^{4}C_{3}\left (\frac{x}{2}-\frac{2}{x} \right )^{3}+^{4}C_{4}\left (\frac{x}{2}-\frac{2}{x} \right )^{4}$
= $\displaystyle \small 1+4\left (\frac{x}{2}-\frac{2}{x} \right )+6\left (\frac{x}{2}-\frac{2}{x} \right )^{2}$$\displaystyle \small +4\left (\frac{x}{2}-\frac{2}{x} \right )^{3}$$\displaystyle \small +\left (\frac{x}{2}-\frac{2}{x} \right )^{4}$
= $\displaystyle \small 1+4\left (\frac{x}{2}-\frac{2}{x} \right )+6[ \left(\frac{x}{2} \right)^{2}+ \left(\frac{2}{x} \right)^{2}$$\displaystyle \small -2 \left(\frac{x}{2} \right) \left(\frac{2}{x} \right) ]$ $\displaystyle \small +4 [ \left(\frac{x}{2} \right)^{3}- \left(\frac{2}{x} \right)^{3}-3 \left(\frac{x}{2} \right)^{2} \left(\frac{2}{x} \right)$$\displaystyle \small +3 \left(\frac{x}{2} \right) \left(\frac{2}{x} \right)^{2} ]$ $\displaystyle \small + [ ^{4}C_{0} \left(\frac{x}{2} \right)^{4}-^{4}C_{1} \left(\frac{x}{2} \right)^{3} \left(\frac{2}{x} \right)$$\displaystyle \small +^{4}C_{2} \left(\frac{x}{2} \right)^{2} \left(\frac{2}{x} \right)^{2}$$\displaystyle \small -^{4}C_{3} \left(\frac{x}{2} \right) \left(\frac{2}{x} \right)^{3}+^{4}C_{4} \left(\frac{2}{x} \right)^{4} ]$
= $\displaystyle \small 1+\left (2x-\frac{8}{x} \right )+6\left[ \frac{x^{2}}{4} + \frac{4}{x^{2}} -2 \right ]$ $\displaystyle \small +4\left [ \frac{x^{3}}{8} - \frac{8}{x^{3}}- \frac{3x}{2} + \frac{6}{x} \right ]$ $\displaystyle \small +\left [ \frac{x^{4}}{16} -4 \frac{x^{2}}{4} +6 -4 \frac{4}{x^{2}} +\frac{16}{x^{4}} \right ]$
= $\displaystyle \small -5-4x+ \frac{x^{2}}{2}+ \frac{x^{3}}{2}+ \frac{x^{4}}{16}+\frac{16}{x}$$\displaystyle \small +\frac{8}{x^{2}}-\frac{32}{x^{3}}+\frac{16}{x^{4}}$
10. Find the expansion of $\displaystyle \small (3x^{2}-2ax+3a^{2})^{3}$ using binomial theorem.
$\displaystyle \small (3x^{2}-2ax+3a^{2})^{3}=((3x^{2}-2ax)+3a^{2})^{3}$
= $\displaystyle \small ^{3}C_{0}(3x^{2}-2ax)^{3}+^{3}C_{1}(3x^{2}-2ax)^{2}(3a^{2})$$\displaystyle \small +^{3}C_{2}(3x^{2}-2ax)(3a^{2})^{2}+^{3}C_{3}(3a^{2})^{3}$
= $\displaystyle \small (3x^{2}-2ax)^{3}+3(3x^{2}-2ax)^{2}(3a^{2})$$\displaystyle \small +3(3x^{2}-2ax)(3a^{2})^{2}+(3a^{2})^{3}$
= $\displaystyle \small (3x^{2})^{3}-(2ax)^{3}-3(3x^{2})^{2}(2ax)$$\displaystyle \small +3(3x^{2})(2ax)^{2}+9a^{2}[(3x^{2})^{2}+(2ax)^{2}$ $\displaystyle \small -2(3x^{2})(2ax)]+(9x^{2}-6ax)(9a^{4})+27a^{6}$
= $\displaystyle \small 27x^{6}-8a^{3}x^{3}-6ax(9x^{4})+9x^{2}(4a^{2}x^{2})$$\displaystyle \small +9a^{2}[9x^{4}+4a^{2}x^{2}-12ax^{3}]$ $\displaystyle \small +81a^{4}x^{2}-54a^{5}x+27a^{6}$
= $\displaystyle \small 27x^{6}-8a^{3}x^{3}-54ax^{5}+36a^{2}x^{4}+81a^{2}x^{4}$ $\displaystyle \small +36a^{4}x^{2}-108a^{3}x^{3}]$ $\displaystyle \small +81a^{4}x^{2}-54a^{5}x+27a^{6}$
= $\displaystyle \small 27x^{6}-54ax^{5}+(36a^{2}x^{4}+81a^{2}x^{4})$$\displaystyle \small +(-8a^{3}x^{3}-108a^{3}x^{3})$ $\displaystyle \small +(36a^{4}x^{2}+81a^{4}x^{2})-54a^{5}x+27a^{6}$
= $\displaystyle \small 27x^{6}-54ax^{5}+117a^{2}x^{4}-116a^{3}x^{3}$ $\displaystyle \small +117a^{4}x^{2}-54a^{5}x+27a^{6}$
We know that,
$\displaystyle \small (a+b)^{n}=^{n}C_{0}a^{n}+^{n}C_{1}a^{n-1}b$$\displaystyle \small +^{n}C_{2}a^{n-2}b^{2}+...$
Given the values of first three terms of the exapnsion.
∴ $\displaystyle \small ^{n}C_{0}a^{n}=729$
$\displaystyle \small ^{n}C_{1}a^{n-1}b=7290$
$\displaystyle \small ^{n}C_{2}a^{n-2}b^{2}=30375$
$\displaystyle \small ^{n}C_{0}a^{n}=729$
⇒ $\displaystyle \small a^{n}=729$ ...(i)
$\displaystyle \small ^{n}C_{1}a^{n-1}b=7290$
⇒ $\displaystyle \small \frac{n!}{1!(n-1)!}a^{n-1}b=7290$
$\displaystyle \small \frac{n(n-1)!}{(n-1)!}a^{n}a^{-1}b=7290$
$\displaystyle \small n \frac{a^{n}}{a}b=7290$
$\displaystyle \small n \frac{729}{a}b=7290$ [from (i)]
$\displaystyle \small n \frac{b}{a}=\frac{7290}{729}$
$\displaystyle \small n \frac{b}{a}=10$ ...(ii)
$\displaystyle \small ^{n}C_{2}a^{n-2}b^{2}=30375$
⇒ $\displaystyle \small \frac{n!}{2!(n-2)!}a^{n}a^{-2}b^{2}=30375$
$\displaystyle \small \frac{n(n-1)(n-2)!}{2(n-2)!}a^{n}a^{-2}b^{2}=30375$
$\displaystyle \small \frac{n(n-1)}{2}\frac{a^{n}}{a^{2}}b^{2}=30375$
$\displaystyle \small \frac{n(n-1)}{2}\frac{729}{a^{2}}b^{2}=30375$
$\displaystyle \small n(n-1)\frac{b^{2}}{a^{2}}=\frac{30375*2}{729}=\frac{250}{3}$ ...(iii)
Divide (iii) by square of (ii),
⇒ $\displaystyle \small \frac{n(n-1)\frac{b^{2}}{a^{2}}}{(n \frac{b}{a})^{2}}=\frac{\frac{250}{3}}{10^2}$
$\displaystyle \small \frac{n(n-1)\frac{b^{2}}{a^{2}}}{n^{2} \frac{b^{2}}{a^{2}}}=\frac{\frac{250}{3}}{100}$
$\displaystyle \small \frac{(n-1)}{n}=\frac{250}{300}$
$\displaystyle \small \frac{n-1}{n}=\frac{5}{6}$
$\displaystyle \small 6n-6=5n$
$\displaystyle \small n=6$
(i) ⇒ $\displaystyle \small a^{6}=729$
$\displaystyle \small 3^{6}=729$
$\displaystyle \small a=3$
(ii) ⇒ $\displaystyle \small n \frac{b}{a}=10$
$\displaystyle \small 6 \frac{b}{3}=10$
$\displaystyle \small b=5$
2. Find $a$ if the coefficients of $x^{2}$ and $x^{3}$ in the expansion of $\displaystyle \small (3+ax)^{9}$ are equal.
general expansion of $\displaystyle \small (3+ax)^{9}$ is,
$\displaystyle \small T_{r+1}=^{9}C_{r}(3)^{9-r}(ax)^{r}$
Put $r=2$ in (i),
⇒ $\displaystyle \small T_{2+1}=^{9}C_{2}(3)^{9-2}(ax)^{2}$
$\displaystyle \small T_{3}=^{9}C_{2}(3)^{7}a^{2}x^{2}$
Put $r=3$ in (i),
⇒ $\displaystyle \small T_{3+1}=^{9}C_{3}(3)^{9-3}(ax)^{3}$
$\displaystyle \small T_{4}=^{9}C_{3}(3)^{6}a^{3}x^{3}$
Given that coefficients of $x^{2}$ and $x^{3}$ are equal.
⇒ $\displaystyle \small ^{9}C_{2}(3)^{7}a^{2}=^{9}C_{3}(3)^{6}a^{3}$
$\displaystyle \small (36)(3)^{7}a^{2}=(84)(3)^{6}a^{3}$
$\displaystyle \small (36)(3)=(84)a$
$\displaystyle \small a=\frac{9}{7}$
3. Find the coefficient of $x^{5}$ in the product $\displaystyle \small (1+2x)^{6}(1-x)^{7}$ using binomial theorem.
$\displaystyle \small (1+2x)^{6}(1-x)^{7}$ =$\displaystyle \small [^{6}C_{0}+^{6}C_{1}(2x)+^{6}C_{2}(2x)^{2}+^{6}C_{3}(2x)^{3}$$\displaystyle \small +^{6}C_{4}(2x)^{4}+^{6}C_{5}(2x)^{5}+^{6}C_{6}(2x)^{6}]$ $\displaystyle \small [^{7}C_{0}-^{7}C_{1}x+^{7}C_{2}x^{2}-^{7}C_{3}x^{3}$$\displaystyle \small +^{7}C_{4}x^{4}-^{7}C_{5}x^{5}+^{7}C_{6}x^{6}-^{7}C_{7}x^{7}]$
= $\displaystyle \small [1+6(2x)+15(2x)^{2}+20(2x)^{3}+15(2x)^{4}$$\displaystyle \small +6(2x)^{5}+(2x)^{6}]$ $\displaystyle \small [1-7x+21x^{2}-35x^{3}+35x^{4}-21x^{5}+7x^{6}$$\displaystyle \small -x^{7}]$
= $\displaystyle \small [1+12x+60x^{2}+160x^{3}+240x^{4}+192x^{5}$$\displaystyle \small +64x^{6}]$ $\displaystyle \small [1-7x+21x^{2}-35x^{3}+35x^{4}-21x^{5}+7x^{6}$$\displaystyle \small -x^{7}]$
coefficient of $x^{5}$ in the product is,
= $\displaystyle \small [-1*21+12*35-60*35+160*21$$\displaystyle \small -240*7+192*1]$
= $\displaystyle \small -21+420-2100+3360-1680+192$$\displaystyle \small =171$
4. If $a$ and $b$ are distinct integers, prove that $a-b$ is a factor of $a^{n}-b^{n}$, whenever n is a positive integer. [Hint: write $a^{n}=(a-b+b)^{n}$ and expand]
Let $\displaystyle \small a^{n}=((a-b)+b)^{n}$ = $\displaystyle \small ^{n}C_{0}(a-b)^{n}+^{n}C_{1}(a-b)^{n-1}b$$\displaystyle \small +^{n}C_{2}(a-b)^{n-2}b^{2}+...+^{n}C_{n-1}(a-b)b^{n-1}$$\displaystyle \small +^{n}C_{n}b^{n}$
$\displaystyle \small a^{n}=(a-b)^{n}+^{n}C_{1}(a-b)^{n-1}b$$\displaystyle \small +^{n}C_{2}(a-b)^{n-2}b^{2}+...+^{n}C_{n-1}(a-b)b^{n-1}$$\displaystyle \small +b^{n}$
$\displaystyle \small a^{n}-b^{n}=(a-b)^{n}+^{n}C_{1}(a-b)^{n-1}b$$\displaystyle \small +^{n}C_{2}(a-b)^{n-2}b^{2}+...+^{n}C_{n-1}(a-b)b^{n-1}$
$\displaystyle \small a^{n}-b^{n}=(a-b)[(a-b)^{n-1}$$\displaystyle \small +^{n}C_{1}(a-b)^{n-2}b+^{n}C_{2}(a-b)^{n-2}b^{3}+...$$\displaystyle \small +^{n}C_{n-1}b^{n-1}]$
which shows $a-b$ is a factor of $a^{n}-b^{n}$
5. Evaluate $\displaystyle \small (\sqrt{3}+\sqrt{2})^{6}-(\sqrt{3}-\sqrt{2})^{6}$
$\displaystyle \small (\sqrt{3}+\sqrt{2})^{6}-(\sqrt{3}-\sqrt{2})^{6}$ = $\displaystyle \small [^{6}C_{0}(\sqrt{3})^{6}+^{6}C_{1}(\sqrt{3})^{5}\sqrt{2}$$\displaystyle \small +^{6}C_{2}(\sqrt{3})^{4}(\sqrt{2})^{2}$ $\displaystyle \small +^{6}C_{3}(\sqrt{3})^{3}(\sqrt{2})^{3}+^{6}C_{4}(\sqrt{3})^{2}(\sqrt{2})^{4}$$\displaystyle \small +^{6}C_{5}\sqrt{3}(\sqrt{2})^{5}+^{6}C_{6}(\sqrt{2})^{6}]$ $\displaystyle \small -[^{6}C_{0}(\sqrt{3})^{6}-^{6}C_{1}(\sqrt{3})^{5}\sqrt{2}+^{6}C_{2}(\sqrt{3})^{4}(\sqrt{2})^{2}$$\displaystyle \small -^{6}C_{3}(\sqrt{3})^{3}(\sqrt{2})^{3}$ $\displaystyle \small +^{6}C_{4}(\sqrt{3})^{2}(\sqrt{2})^{4}$$\displaystyle \small -^{6}C_{5}\sqrt{3}(\sqrt{2})^{5}+^{6}C_{6}(\sqrt{2})^{6}]$
= $\displaystyle \small 2(^{6}C_{1}(\sqrt{3})^{5}\sqrt{2})+2(^{6}C_{3}(\sqrt{3})^{3}(\sqrt{2})^{3})$$\displaystyle \small +2(^{6}C_{5}\sqrt{3}(\sqrt{2})^{5})$
= $\displaystyle \small 2(6)(\sqrt{3})^{5}\sqrt{2}+2(20)(\sqrt{3})^{3}(\sqrt{2})^{3}$$\displaystyle \small +2(6)\sqrt{3}(\sqrt{2})^{5}$
= $\displaystyle \small 2(6)(9)\sqrt{3}\sqrt{2}+2(20)(3)\sqrt{3}(2)\sqrt{2}$$\displaystyle \small +2(6)\sqrt{3}(4)\sqrt{2}$
= $\displaystyle \small 108\sqrt{6}+240\sqrt{6}+48\sqrt{6}=396\sqrt{6}$
6. Find the value of $\displaystyle \small (a^{2}+\sqrt{a^{2}-1})^{4}+(a^{2}-\sqrt{a^{2}-1})^{4}$
$\displaystyle \small (a^{2}+\sqrt{a^{2}-1})^{4}+(a^{2}-\sqrt{a^{2}-1})^{4}$= $\displaystyle \small [^{4}C_{0}(a^{2})^{4}+^{4}C_{1}(a^{2})^{3}(\sqrt{a^{2}-1})$$\displaystyle \small +^{4}C_{2}(a^{2})^{2}(\sqrt{a^{2}-1})^{2}$ $\displaystyle \small +^{4}C_{3}a^{2}(\sqrt{a^{2}-1})^{3}+^{4}C_{4}(\sqrt{a^{2}-1})^{4}]$$\displaystyle \small +[^{4}C_{0}(a^{2})^{4}-^{4}C_{1}(a^{2})^{3}(\sqrt{a^{2}-1})$ $\displaystyle \small +^{4}C_{2}(a^{2})^{2}(\sqrt{a^{2}-1})^{2}-^{4}C_{3}a^{2}(\sqrt{a^{2}-1})^{3}$$\displaystyle \small +^{4}C_{4}(\sqrt{a^{2}-1})^{4}]$
= $\displaystyle \small 2(^{4}C_{0}(a^{2})^{4})+2(^{4}C_{2}(a^{2})^{2}(\sqrt{a^{2}-1})^{2})$$\displaystyle \small +2(^{4}C_{4}(\sqrt{a^{2}-1})^{4})$
= $\displaystyle \small 2(a^{2})^{4}+2(6)(a^{2})^{2}(\sqrt{a^{2}-1})^{2}$$\displaystyle \small +2(\sqrt{a^{2}-1})^{4}$
= $\displaystyle \small 2a^{8}+12a^{4}(a^{2}-1)+2(a^{2}-1)^{2}$
= $\displaystyle \small 2[a^{8}+6a^{4}(a^{2}-1)+(a^{2}-1)^{2}]$
= $\displaystyle \small 2[a^{8}+6a^{6}-6a^{4}+a^{4}+1-2a^{2}]$
= $\displaystyle \small 2[a^{8}+6a^{6}-5a^{4}-2a^{2}+1]$
7. Find an approximation of $(0.99)^{5}$ using the first three terms of its expansion.
$\displaystyle \small (0.99)^{5}=(1-0.01)^{5}$
= $\displaystyle \small ^{5}C_{0}-^{5}C_{1}(0.01)+^{5}C_{2}(0.01)^{2}+...$
= $\displaystyle \small 1-5(0.01)+10(0.01)^{2}+...$
= $\displaystyle \small 1-0.05+0.001+...=0.951$
8. Find n if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of $\displaystyle \small \left ( \sqrt[4]{2}+\frac{1}{\sqrt[4]{3}} \right )^{n}$ is $\displaystyle \small \sqrt{6}:1$ .
We know that, $\displaystyle \small T_{r+1}=^{n}C_{r}a^{n-r}b^{r}$
fifth term from the beginning is when $r=4$
fifth term from the end is when $r=n-4$
∴ $\displaystyle \small T_{4+1}=^{n}C_{4} \left (\sqrt[4]{2} \right )^{n-4} \left ( \frac{1}{\sqrt[4]{3}} \right )^{4}$
$\displaystyle \small T_{n-4+1}=^{n}C_{n-4} \left (\sqrt[4]{2} \right )^{n-(n-4)} \left ( \frac{1}{\sqrt[4]{3}} \right )^{n-4}$
Given that, $\displaystyle \small T_{4+1}:T_{n-4+1}=\sqrt{6}:1$
$\displaystyle \small \frac{^{n}C_{4} \left (\sqrt[4]{2} \right )^{n-4} \left ( \frac{1}{\sqrt[4]{3}} \right )^{4}}{^{n}C_{n-4} \left (\sqrt[4]{2} \right )^{4} \left ( \frac{1}{\sqrt[4]{3}} \right )^{n-4}}=\frac{\sqrt{6}}{1}$
$\displaystyle \small \frac{ \left (\sqrt[4]{2} \right )^{n-4} \left ( \frac{1}{\sqrt[4]{3}} \right )^{4}}{ \left (\sqrt[4]{2} \right )^{4} \left ( \frac{1}{\sqrt[4]{3}} \right )^{n-4}}=\sqrt{6}$ [$^{n}C_{4}=^{n}C_{n-4}$]
$\displaystyle \small \frac{ \left (\sqrt[4]{2} \right )^{n-4} \left ( \sqrt[4]{3} \right )^{-4}}{ \left (\sqrt[4]{2} \right )^{4} \left (\sqrt[4]{3} \right )^{-n+4}}=\sqrt{6}$
$\displaystyle \small \left (\sqrt[4]{2} \right )^{n-4-4} \left ( \sqrt[4]{3} \right )^{-4+n-4}=\sqrt{6}$
$\displaystyle \small \left (\sqrt[4]{2} \right )^{n-8} \left ( \sqrt[4]{3} \right )^{n-8}=\sqrt{6}$
$\displaystyle \small (2)^\frac{n-8}{4}(3)^\frac{n-8}{4}=\sqrt{6}$
$\displaystyle \small (6)^\frac{n-8}{4}=(6)^\frac{1}{2}$
$\displaystyle \small \frac{n-8}{4}=\frac{1}{2}$
$\displaystyle \small n-8=2$
$\displaystyle \small n=10$
9. Expand using Binomial theorem $\displaystyle \small \left ( 1+\frac{x}{2}-\frac{2}{x} \right )^{4}$
$\displaystyle \small \left ( 1+\frac{x}{2}-\frac{2}{x} \right )^{4}=\left ( 1+\left (\frac{x}{2}-\frac{2}{x} \right ) \right )^{4}$
= $\displaystyle \small ^{4}C_{0}+^{4}C_{1}\left (\frac{x}{2}-\frac{2}{x} \right )+^{4}C_{2}\left (\frac{x}{2}-\frac{2}{x} \right )^{2}$$\displaystyle \small +^{4}C_{3}\left (\frac{x}{2}-\frac{2}{x} \right )^{3}+^{4}C_{4}\left (\frac{x}{2}-\frac{2}{x} \right )^{4}$
= $\displaystyle \small 1+4\left (\frac{x}{2}-\frac{2}{x} \right )+6\left (\frac{x}{2}-\frac{2}{x} \right )^{2}$$\displaystyle \small +4\left (\frac{x}{2}-\frac{2}{x} \right )^{3}$$\displaystyle \small +\left (\frac{x}{2}-\frac{2}{x} \right )^{4}$
= $\displaystyle \small 1+4\left (\frac{x}{2}-\frac{2}{x} \right )+6[ \left(\frac{x}{2} \right)^{2}+ \left(\frac{2}{x} \right)^{2}$$\displaystyle \small -2 \left(\frac{x}{2} \right) \left(\frac{2}{x} \right) ]$ $\displaystyle \small +4 [ \left(\frac{x}{2} \right)^{3}- \left(\frac{2}{x} \right)^{3}-3 \left(\frac{x}{2} \right)^{2} \left(\frac{2}{x} \right)$$\displaystyle \small +3 \left(\frac{x}{2} \right) \left(\frac{2}{x} \right)^{2} ]$ $\displaystyle \small + [ ^{4}C_{0} \left(\frac{x}{2} \right)^{4}-^{4}C_{1} \left(\frac{x}{2} \right)^{3} \left(\frac{2}{x} \right)$$\displaystyle \small +^{4}C_{2} \left(\frac{x}{2} \right)^{2} \left(\frac{2}{x} \right)^{2}$$\displaystyle \small -^{4}C_{3} \left(\frac{x}{2} \right) \left(\frac{2}{x} \right)^{3}+^{4}C_{4} \left(\frac{2}{x} \right)^{4} ]$
= $\displaystyle \small 1+\left (2x-\frac{8}{x} \right )+6\left[ \frac{x^{2}}{4} + \frac{4}{x^{2}} -2 \right ]$ $\displaystyle \small +4\left [ \frac{x^{3}}{8} - \frac{8}{x^{3}}- \frac{3x}{2} + \frac{6}{x} \right ]$ $\displaystyle \small +\left [ \frac{x^{4}}{16} -4 \frac{x^{2}}{4} +6 -4 \frac{4}{x^{2}} +\frac{16}{x^{4}} \right ]$
= $\displaystyle \small -5-4x+ \frac{x^{2}}{2}+ \frac{x^{3}}{2}+ \frac{x^{4}}{16}+\frac{16}{x}$$\displaystyle \small +\frac{8}{x^{2}}-\frac{32}{x^{3}}+\frac{16}{x^{4}}$
10. Find the expansion of $\displaystyle \small (3x^{2}-2ax+3a^{2})^{3}$ using binomial theorem.
$\displaystyle \small (3x^{2}-2ax+3a^{2})^{3}=((3x^{2}-2ax)+3a^{2})^{3}$
= $\displaystyle \small ^{3}C_{0}(3x^{2}-2ax)^{3}+^{3}C_{1}(3x^{2}-2ax)^{2}(3a^{2})$$\displaystyle \small +^{3}C_{2}(3x^{2}-2ax)(3a^{2})^{2}+^{3}C_{3}(3a^{2})^{3}$
= $\displaystyle \small (3x^{2}-2ax)^{3}+3(3x^{2}-2ax)^{2}(3a^{2})$$\displaystyle \small +3(3x^{2}-2ax)(3a^{2})^{2}+(3a^{2})^{3}$
= $\displaystyle \small (3x^{2})^{3}-(2ax)^{3}-3(3x^{2})^{2}(2ax)$$\displaystyle \small +3(3x^{2})(2ax)^{2}+9a^{2}[(3x^{2})^{2}+(2ax)^{2}$ $\displaystyle \small -2(3x^{2})(2ax)]+(9x^{2}-6ax)(9a^{4})+27a^{6}$
= $\displaystyle \small 27x^{6}-8a^{3}x^{3}-6ax(9x^{4})+9x^{2}(4a^{2}x^{2})$$\displaystyle \small +9a^{2}[9x^{4}+4a^{2}x^{2}-12ax^{3}]$ $\displaystyle \small +81a^{4}x^{2}-54a^{5}x+27a^{6}$
= $\displaystyle \small 27x^{6}-8a^{3}x^{3}-54ax^{5}+36a^{2}x^{4}+81a^{2}x^{4}$ $\displaystyle \small +36a^{4}x^{2}-108a^{3}x^{3}]$ $\displaystyle \small +81a^{4}x^{2}-54a^{5}x+27a^{6}$
= $\displaystyle \small 27x^{6}-54ax^{5}+(36a^{2}x^{4}+81a^{2}x^{4})$$\displaystyle \small +(-8a^{3}x^{3}-108a^{3}x^{3})$ $\displaystyle \small +(36a^{4}x^{2}+81a^{4}x^{2})-54a^{5}x+27a^{6}$
= $\displaystyle \small 27x^{6}-54ax^{5}+117a^{2}x^{4}-116a^{3}x^{3}$ $\displaystyle \small +117a^{4}x^{2}-54a^{5}x+27a^{6}$
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