QJEEM24Ma76c0: JEE Main 2024 Mathematics - Class 11 JEEM24Ma76c0 Solution: In the expansion of , the sum of the coefficients of and is…

$\begin{aligned}& (1+x)\left(1-x^2\right)\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}\right)^5 \\& =(1+x)\left(1-x^2\right)\left(\left(1+\frac{1}{x}\right)^3\right)^5 \\& =\frac{(1+x)^2(1-x)(1+x)^{15}}{x^{15}} \\& =\frac{(1+x)^{17}-x(1+x)^{17}}{x^{15}}\end{aligned}$

$=\operatorname{coeff}\left(\mathrm{x}^3\right)$ in the expansion $\approx \operatorname{coeff}\left(\mathrm{x}^{18}\right)$ in

$\begin{aligned}& (1+x)^{17}-x(1+x)^{17} \\& =0-1 \\& =-1\end{aligned}$

$\operatorname{coeff}\left(\mathrm{x}^{-13}\right)$ in the expansion $\approx \operatorname{coeff}\left(\mathrm{x}^2\right)$ in

$\begin{aligned}& (1+x)^{17}-x(1+x)^{17} \\& =\left(\begin{array}{c}17 \\2\end{array}\right)-\left(\begin{array}{c}17 \\1\end{array}\right) \\& =17 \times 8-17 \\& =17 \times 7 \\& =119\end{aligned}$

Hence Answer $=119-1=118$