Question

Find the coefficient of x^12 in the exapansion of (x^2+1/x^2) ^10

Answer 1

$\large\underline{\sf{Solution-}}$

Given expansion is

$\rm :\longmapsto\: {\bigg[ {x}^{2} + \dfrac{1}{ {x}^{2} } \bigg]}^{10}$

We know,

General Term in the expansion of

$\boxed{ \tt{ \: {(x + y)}^{n} \: is \: T_{r + 1} = \: ^{n}C_{r} {x}^{n - r} {y}^{r} \: }}$

So, General Term in the expansion of

$\rm :\longmapsto\: {\bigg[ {x}^{2} + \dfrac{1}{ {x}^{2} } \bigg]}^{10}$

is

$\rm :\longmapsto\:T_{r + 1} = \: ^{10}C_{r} {\bigg[ {x}^{2} \bigg]}^{10 - r} {\bigg[\dfrac{1}{ {x}^{2} } \bigg]}^{r}$

$\rm :\longmapsto\:T_{r + 1} = \: ^{10}C_{r} {\bigg[ {x}^{} \bigg]}^{20 - 2r} {\bigg[\dfrac{1}{ {x}^{2r} } \bigg]}^{}$

$\rm :\longmapsto\:T_{r + 1} = \: ^{10}C_{r} {\bigg[ {x}^{} \bigg]}^{20 - 2r - 2r}$

$\rm :\longmapsto\:T_{r + 1} = \: ^{10}C_{r} {\bigg[ {x}^{} \bigg]}^{20 - 4r}$

Since, we have to find coefficient of x^12, so substitute

$\red{\rm :\longmapsto\:20 - 4r = 12}$

$\red{\rm :\longmapsto\: - 4r = 12 - 20}$

$\red{\rm :\longmapsto\: - 4r = -8 }$

$\red{\bf\implies \:\: r =2 }$

Hence,

$\rm :\longmapsto\:T_{2 + 1} = \: ^{10}C_{2} {\bigg[ {x}^{} \bigg]}^{20 - 4 \times 2}$

$\rm :\longmapsto\:T_{3} = \: \dfrac{10 \times 9}{2 \times 1} {\bigg[ {x}^{} \bigg]}^{20 - 8}$

$\rm :\longmapsto\:T_{3} = \: 45{\bigg[ {x}^{} \bigg]}^{12}$

$\purple{\bf\implies \:\boxed{ \tt{ \: coefficient \: of \: {x}^{12} = 45 \: \: }}}$

More to know

$\boxed{ \tt{ \: ^{n}C_{x} = \: ^{n}C_{y}\bf\implies \:x = y \: \: or \: \: n = x + y \: }}$

$\boxed{ \tt{ \: ^{n}C_{r} = \frac{n - 1}{r} \: ^{n - 1}C_{r - 1} \: \: }}$

$\boxed{ \tt{ \: ^{n}C_{r} \: + \: ^{n}C_{r - 1} \: = \: ^{n + 1}C_{r} \: }}$

$\boxed{ \tt{ \: \frac{ ^{n}C_{r}}{ ^{n}C_{r - 1}} = \frac{n - r + 1}{r} \: \: }}$