In the expansion of (1+x)left(1-x^2right)left | vedclass

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Question

$\begin{aligned} & (1+x)\left(1-x^2ight)\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}ight)^5 \ & =(1+x)\left(1-x^2ight)\left(\left(1+

Vedclass · Accepted Answer

$118$

Vedclass · Answer

$\begin{aligned} & (1+x)\left(1-x^2ight)\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}ight)^5 \ & =(1+x)\left(1-x^2ight)\left(\left(1+\frac{1}{x}ight)^3ight)^5 \ & =\frac{(1+x)^2(1-x)(1+x)^{15}}{x^{15}} \ & =\frac{(1+x)^{17}-x(1+x)^{17}}{x^{15}} \ & =\operatorname{coeff}\left(x^3ight) 	ext { in the expansion } \approx \operatorname{coeff}\left(x^{18}ight) 	ext { in } \ & (1+x)^{17}-x(1+x)^{17} \ & =0-1 \ & =-1 \ & 	ext { coeff }\left(x^{-13}ight) 	ext { in the expansion } \approx \operatorname{coeff}\left(x^2ight) \quad 	ext { in }\end{aligned}$

$(1+x)^{17}-x(1+x)^{17}$

$=0-1$

$=-1$

$\operatorname{coeff}\left(\mathrm{x}^{-13}ight)$ in the expansion $\approx \operatorname{coeff}\left(\mathrm{x}^2ight)$ in

$(1+x)^{17}-x(1+x)^{17}$

$=\left(\begin{array}{c}17 \ 2\end{array}ight)-\left(\begin{array}{c}17 \ 1\end{array}ight)$

$=17 	imes 8-17$

$=17 	imes 7$

$=119$

Hence Answer $=119-1=118$

In the expansion of $(1+x)\left(1-x^2\right)\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}\right)^5, x \neq 0$, the sum of the coefficient of $x^3$ and $x^{-13}$ is equal to

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