The sum 1+1+3/2!+1+3+5/3!+1+3+5+7/4!+ldots upto ∞ terms, is equal to

Mathematics · Sequence and Series

JEE Main 2025 — 3 April, Evening Shift — Question 30

The sum $1+\frac{1+3}{2!}+\frac{1+3+5}{3!}+\frac{1+3+5+7}{4!}+\ldots$ upto $\infty$ terms, is equal to

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

$\sum_{r=1}^{\infty} T_{r}=\sum_{r=1}^{\infty} \frac{1}{(r-2)!}+\frac{1}{(r-1)!}=e+e=2 e$

Answer key and solution verified before publishing.

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.