The sum upto terms, is equal to [2025]

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Solution

Q.
The sum $1 + \frac{1 + 3}{2!} + \frac{1 + 3 + 5}{3!} + \frac{1 + 3 + 5 + 7}{4!} + . . .$ upto $\infty$ terms, is equal to [2025]

1 6e

2 3e

3 2e

4 4e

Ans.
(3)

Let $S = 1 + \frac{1 + 3}{2!} + \frac{1 + 3 + 5}{3!} + . . .$

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

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

$= 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + . . . . + 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + . . . = e + e = 2 e$ .

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