JEE Main 2025 Apr 3 Shift 2, Mathematics Q20: Summation of Series

The $r$-th term (starting from $r=1$) has numerator $1+3+5+\cdots +(2r-1)=r^2$ and denominator $r!$. So

$S=\sum _{r=1}^{\infty }\frac{r^2}{r!}$.

Use $r^2=r(r-1)+r$. Then $\frac{r^2}{r!}=\frac{r(r-1)}{r!}+\frac{r}{r!}=\frac{1}{(r-2)!}+\frac{1}{(r-1)!}$ (with the convention that terms for $r=1$ contribute only the second piece).

$\sum _{r=2}^{\infty }\frac{1}{(r-2)!}=e$ and $\sum _{r=1}^{\infty }\frac{1}{(r-1)!}=e$.

$\therefore S=e+e=2e$.