First slide

Sigma operation series

Question

$The sum of the series \sum_{n = 1}^{\infty} \frac{n^{2} + 6 n + 10}{(2 n + 1)!} is equal to$

Moderate

A

$\frac{41}{8} e + \frac{19}{8} e^{- 1} + 10$
B

$- \frac{41}{8} e + \frac{19}{8} e^{- 1} - 10$
C

$\frac{41}{8} e - \frac{19}{8} e^{- 1} - 10$
D

$\frac{41}{8} e + \frac{19}{8} e^{- 1} - 10$

Solution

$\begin{array}{l} \sum_{n = 1}^{\infty} \frac{n^{2} + 6 n + 10}{(2 n + 1)!} Let 2 n + 1 = λ \Rightarrow n = \frac{λ - 1}{2} \\ \frac{n^{2} + 6 n + 10}{(2 n + 1)!} = \frac{{(\frac{λ - 1}{2})}^{2} + 6 (\frac{λ - 1}{2}) + 10}{λ!} = \frac{λ^{2} + 10 λ + 29}{4 (λ!)} \\ Also λ = 3, 5, 7, 9, 11, \dots \dots \\ ∴ Required = \frac{1}{4} \sum \frac{λ (λ - 1) + 11 λ + 29}{λ!} \\ λ = 3, 5, 7, \dots \dots \dots . \end{array}$
$\begin{array}{l} = \frac{1}{4} (\sum_{λ = 3, 5, 7, \dots} (\frac{1}{(λ - 2)!} + \frac{11}{(λ - 1)!} + \frac{29}{λ!})) \\ = \frac{1}{4} ((\frac{1}{4} + \frac{1}{⌊ 3} + \frac{1}{2} + \dots .) + 11 (\frac{1}{⌊ 2} + \frac{1}{4} + \dots . . +) + 29 (\frac{1}{⌊ 3} + \frac{1}{15} + \frac{1}{17} + \dots \dots)) \\ = \frac{1}{4} (\frac{e - \frac{1}{e}}{2}) + 11 (\frac{e + \frac{1}{e} - 2}{2}) + 29 (\frac{e - \frac{1}{e} - 2}{2}) = \frac{1}{8} (41 e - \frac{19}{c} - 80) \end{array}$

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