JEE Main 2024 — Binomial Theorem Question with Solution

$\begin{aligned} & \mathrm{f}(\mathrm{x})=1+\frac{(1+\mathrm{x})}{1 !}+\frac{(1+\mathrm{x})^2}{2 !}+\frac{(1+\mathrm{x})^3}{3 !}+\ldots . . \\ & \frac{\mathrm{e}^{(1+\mathrm{x})}}{1+\mathrm{x}}=\frac{1}{1+\mathrm{x}}+1+\frac{(1+\mathrm{x})}{2 !}+\frac{(1+\mathrm{x})^2}{3 !}+\frac{(1+\mathrm{x})^2}{4 !} \\ & \text { coef } \mathrm{x}^2 \text { in RHS : } 1+\frac{{ }^2 \mathrm{C}_2}{3}+\frac{{ }^3 \mathrm{C}_2}{4}+\ldots=\mathrm{a} \end{aligned}$ coeff. $x^2$ in L.H.S. $e\left(1+x+\frac{x^2}{2!}\right)\dots \left(1-x+\frac{x^2}{2!}\dots \dots \right)$ is $\mathrm{e}-\mathrm{e}+\frac{\mathrm{e}}{2!}=\mathrm{a}$ $\begin{aligned} & \mathrm{b}=1+\frac{2}{1 !}+\frac{2^2}{2 !}+\frac{2^3}{3 !}+\ldots \ldots=\mathrm{e}^2 \\ & \frac{2 \mathrm{~b}}{\mathrm{a}^2}=8 \end{aligned}$