The sum sum limits_{n=1}^{infty} frac{2 n^2+3 | vedclass

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Question

$\sum \limits_{ n =1}^{\infty} \frac{2 n ^2+3 n +4}{(2 n ) !}$

$\frac{1}{2} \sum \limits_{ n =1}^{\infty} \frac{2 n (2 n -1)+8 n +8}{(2 n ) !}$

Vedclass · Accepted Answer

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

Vedclass · Answer

$\sum \limits_{ n =1}^{\infty} \frac{2 n ^2+3 n +4}{(2 n ) !}$

$\frac{1}{2} \sum \limits_{ n =1}^{\infty} \frac{2 n (2 n -1)+8 n +8}{(2 n ) !}$

$\frac{1}{2} \sum \limits_{ n =1}^{\infty} \frac{1}{(2 n -2) !}+2 \sum \limits_{ n =1}^{\infty} \frac{1}{(2 n -1) !}+4 \sum \limits_{ n =1}^{\infty} \frac{1}{(2 n ) !}$

$e =1+1+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\ldots \ldots$

$e ^{-1}=1-1+\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}+\ldots \ldots$

$\left( e +\frac{1}{ e }ight)=2\left(1+\frac{1}{2 !}+\frac{1}{4 !}+\ldots \ldots .ight)$

$e -\frac{1}{ e }=\left(1+\frac{1}{3 !}+\frac{1}{5 !}+\ldots . .ight)$

$	ext { Now }$

$\frac{1}{2}\left(\sum \limits_{ n =1}^{\infty} \frac{1}{(2 n -2) !}ight)+2 \sum \limits_{ n =1}^{\infty} \frac{1}{(2 n -1) !}+4 \sum \limits_{ n =1}^{\infty} \frac{1}{(2 n ) !}$

$=\frac{1}{2}\left[\frac{ e +\frac{1}{ e }}{2}ight]+2\left[\frac{ e -\frac{1}{ e }}{2}ight]+4\left(\frac{ e +\frac{1}{ e }-2}{2}ight)$

$=\frac{\left( e +\frac{1}{ e }ight)}{4}+ e -\frac{1}{ e }+2 e +\frac{2}{ e }-4$

$=\frac{13}{4} e +\frac{5}{4 e }-4$

The sum $\sum \limits_{n=1}^{\infty} \frac{2 n^2+3 n+4}{(2 n) !}$ is equal to :

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