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

Correct answer: D. 2 e aligned & S=1+(1+3)/(2!)+(1+3+5)/(3!)+ & =∑_ r=1 ^ ∞ r^2 r! & =∑_ r=1 ^ ∞ (r-1+1) (r-1)! =∑_ r=2 ^ ∞ 1 (r-2)! +∑_ r=1 ^ ∞ 1 (r-1)! & =2 e aligned

JEE Main 20253 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 3 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Sequences and Series

JEE Main 2025 · 3 April · Evening Shift · Mathematics · Sequences and Series

The sum 1+(1+3)/(2!)+(1+3+5)/(3!)+(1+3+5+7)/(4!)+ upto ∞ terms, is...

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

Options

A $6 e$
B $4 e$
C $3 e$
D $2 e$

Correct answer

D. $2 e$

Step-by-step solution

\begin{aligned} & \mathrm{S}=1+\frac{1+3}{2!}+\frac{1+3+5}{3!}+\ldots \\ & =\sum_{\mathrm{r}=1}^{\infty} \frac{\mathrm{r}^2}{\mathrm{r}!} \\ & =\sum_{\mathrm{r}=1}^{\infty} \frac{(\mathrm{r}-1+1)}{(\mathrm{r}-1)!}=\sum_{\mathrm{r}=2}^{\infty} \frac{1}{(\mathrm{r}-2)!}+\sum_{\mathrm{r}=1}^{\infty} \frac{1}{(\mathrm{r}-1)!} \\ & =2 \mathrm{e}\end{aligned}

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