JEE Main 2025 — Sequences And Series Question with Solution

The series given is:

$1+\frac{1+3}{2!}+\frac{1+3+5}{3!}+\frac{1+3+5+7}{4!}+\dots$

In general, the $r$-th term of the series, denoted as $T_r$, takes the form:

$T_r=\frac{r^2}{r!}$

To simplify $T_r$, we write it in terms of factorials:

$T_r=\frac{r}{(r-1)!}=\frac{(r-1)+1}{(r-1)!}=\frac{1}{(r-2)!}+\frac{1}{(r-1)!}$

Thus, the series can be expressed as:

$\sum _{r=1}^{\infty }{T}_r=\sum _{r=1}^{\infty }\left(\frac{1}{(r-2)!}+\frac{1}{(r-1)!}\right)$

This breaks into two parts:

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

Each part is a well-known series. Specifically:

$\sum _{r=1}^{\infty }\frac{1}{(r-2)!}=e$, starting from the term where $r-2=0$, which aligns with the expansion of $e$.

$\sum _{r=1}^{\infty }\frac{1}{(r-1)!}=e$, for the terms starting where $r-1=0$.

Consequently, the sum of the entire series is:

$e+e=2e$