The sum of the series ∑n=1∞n2+6n+10(2n+1)! is equal to

∑n=1∞n2+6n+10(2n+1)! Let 2n+1=λ ⇒n=λ-12n2+6n+10(2n+1)!=λ-122+6λ-12+10λ!=λ2+10λ+294(λ!) Also λ=3,5,7,9,11,……∴ Required =14∑λ(λ-1)+11λ+29λ!λ=3,5,7,……….=14∑λ=3,5,7,…1(λ-2)!+11(λ-1)!+29λ!=1414+1⌊3+12+….+111⌊2+14+…..++291⌊3+115+117+……=14e-1e2+11e+1e-22+29e-1e-22 =1841e-19c-80

The sum of the series ∑n=1∞n2+6n+10(2n+1)! is equal to

∑n=1∞n2+6n+10(2n+1)! Let 2n+1=λ ⇒n=λ-12n2+6n+10(2n+1)!=λ-122+6λ-12+10λ!=λ2+10λ+294(λ!) Also λ=3,5,7,9,11,……∴ Required =14∑λ(λ-1)+11λ+29λ!λ=3,5,7,……….=14∑λ=3,5,7,…1(λ-2)!+11(λ-1)!+29λ!=1414+1⌊3+12+….+111⌊2+14+…..++291⌊3+115+117+……=14e-1e2+11e+1e-22+29e-1e-22 =1841e-19c-80

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

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

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

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

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

answer is C.

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Detailed Solution

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

$\begin{array}{l} = \frac{1}{4} (\sum_{λ = 3, 5, 7, \dots} (\frac{1}{(λ - 2)!} + \frac{11}{(λ - 1)!} + \frac{29}{λ!})) \\ = \frac{1}{4} ((\frac{1}{4} + \frac{1}{⌊ 3} + \frac{1}{2} + \dots .) + 11 (\frac{1}{⌊ 2} + \frac{1}{4} + \dots . . +) + 29 (\frac{1}{⌊ 3} + \frac{1}{15} + \frac{1}{17} + \dots \dots)) \\ = \frac{1}{4} (\frac{e - \frac{1}{e}}{2}) + 11 (\frac{e + \frac{1}{e} - 2}{2}) + 29 (\frac{e - \frac{1}{e} - 2}{2}) = \frac{1}{8} (41 e - \frac{19}{c} - 80) \end{array}$