JEE Main 2024 — Binomial Theorem Question with Solution

Given,

$\left(1+x\right)\left(1-x^2\right){\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}\right)}^5$

$={\left(1+x\right)}^2\left(1-x\right){\left(\frac{x^3+3x^2+3x+1}{x^3}\right)}^5$

$={\left(1+x\right)}^2\left(1-x\right){\left(\frac{{\left(1+x\right)}^3}{x^3}\right)}^5$

$=\frac{{\left(1+x\right)}^{17}\left(1-x\right)}{x^{15}}$

$=\frac{{\left(1+x\right)}^{17}}{x^{15}}-\frac{{\left(1+x\right)}^{17}}{x^{14}}$

Now for coefficient of $x^3$, we will find coefficient of $x^{18}$ in expansion of ${\left(1+x\right)}^{17}$ which is not possible and $x^{17}$ in expansion of $-{\left(1+x\right)}^{17}$ which will be $-{}^{17}C_{17}=-1$

And for coefficient of $x^{-13}$ we will coefficient of $x^2$ in expansion of ${\left(1+x\right)}^{17}$ which will be ${}^{17}C_2$ and coefficient of $x$ in expansion of $-{\left(1+x\right)}^{17}$ which will be $-{}^{17}C_1$, so the coefficient of $x^{-13}$ will be ${}^{17}C_2-{}^{17}C_1=136-17=119$

Hence, the sum will be $119-1=118$