JEE Main 2024 — Binomial Theorem Question with Solution

$$\begin{array}{rl}& a=1+\frac{{}^2{C}_2}{3!}+\frac{{}^3{C}_2}{4!}+\frac{{}^4{C}_2}{5!}+\dots \\ & b=1+\frac{{}^1{C}_0+{}^1{C}_1}{1!}+\frac{{}^2{C}_0+{}^2{C}_1+{}^2{C}_2}{2!}+\dots \\ & b=1+\frac{2}{1!}+\frac{2^2}{2!}+\frac{2}{3!}+\dots =e^2\end{array}$$

Using $$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x}{3!}+\dots$$

$$\begin{array}{rl}a=& 1+\sum _{r=2}^{\infty }\frac{{}^r{C}_2}{(r+1)!}=1+\sum _{r=2}\frac{r(r-1)}{2(r+1)!}\\ & =1+\frac{1}{2}\sum _{r=2}^{\infty }\frac{(r+1)r-2r}{(r+1)!}\\ & =1+\frac{1}{2}\sum _{r=2}^{\infty }\frac{1}{(r-1)!}-\frac{1}{2}\sum _{r=2}\frac{2r}{(r+1)!}\\ & =1+\frac{1}{2}\left(\frac{1}{1!}+\frac{1}{2!}+\dots \right)-\sum _{r=2}^{\infty }\frac{(r+1)-1}{(r+1)!}\\ & =1+\frac{1}{2}(e-1)-\sum _{r=2}^{\infty }\frac{1}{r!}+\sum _{r=2}\frac{1}{(r+1)!}\\ & =1+\frac{1}{2}(e-1)-\left(e-\frac{1}{1!}-\frac{1}{0!}\right)+\left(e-\frac{1}{1!}-\frac{1}{0!}-\frac{1}{2!}\right)\\ & =1+\frac{e}{2}-\frac{1}{2}-e+2+e-2-\frac{1}{2}=\frac{e}{2}\\ \Rightarrow & \frac{2b}{a^2}=\frac{2}{\frac{e^2}{4}}8\end{array}$$