The sum n = 1 2 n 2 + 3 n + 4 ( 2 n ) is equal to [2023]

Question

The sum $\sum_{n = 1}^{\infty} \frac{2 n^{2} + 3 n + 4}{(2 n)!}$ is equal to [2023]

Smart Achievers · Accepted Answer

(3)

$\sum_{n = 1}^{\infty} \frac{2 n^{2} + 3 n + 4}{(2 n)!}$

$= \frac{1}{2} \sum_{n = 1}^{\infty} \frac{2 n (2 n - 1) + 8 n + 8}{(2 n)!}$

$= \frac{1}{2} \sum_{n = 1}^{\infty} \frac{1}{(2 n - 2)!} + 2 \sum_{n = 1}^{\infty} \frac{1}{(2 n - 1)!} + 4 \sum_{n = 1}^{\infty} \frac{1}{(2 n)!}$

$e = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \dots$

$e^{- 1} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots$

$(e + \frac{1}{e}) = 2 (1 + \frac{1}{2!} + \frac{1}{4!} + \dots)$

$e - \frac{1}{e} = 2 (1 + \frac{1}{3!} + \frac{1}{5!} + \dots)$

$Now, \frac{1}{2} (\sum_{n = 1}^{\infty} \frac{1}{(2 n - 2)!}) + 2 \sum_{n = 1}^{\infty} \frac{1}{(2 n - 1)!} + 4 \sum_{n = 1}^{\infty} \frac{1}{(2 n)!}$

$= \frac{1}{2} [\frac{e + \frac{1}{e}}{2}] + 2 [\frac{e - \frac{1}{e}}{2}] + 4 [\frac{e + \frac{1}{e} - 2}{2}]$

$= \frac{(e + \frac{1}{e})}{4} + e - \frac{1}{e} + 2 e + \frac{2}{e} - 4$

$= \frac{e}{4} + \frac{1}{4 e} + 3 e + \frac{1}{e} - 4 = \frac{e + 12 e}{4} + \frac{1 + 4}{4 e} - 4$

$= \frac{13 e}{4} + \frac{5}{4 e} - 4$

Solution

Q.
The sum $\sum_{n = 1}^{\infty} \frac{2 n^{2} + 3 n + 4}{(2 n)!}$ is equal to [2023]

1 $\frac{11 e}{2} + \frac{7}{2 e} - 4$

2 $\frac{11 e}{2} + \frac{7}{2 e}$

3 $\frac{13 e}{4} + \frac{5}{4 e} - 4$

4 $\frac{13 e}{4} + \frac{5}{4 e}$

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Solution

Q. The sum ∑n=1∞2n2+3n+4(2n)! is equal to [2023]

1 11e2+72e-4

2 11e2+72e

3 13e4+54e-4