Let quad sum limits_{n=0}^{infty} frac{n^3((2 | vedclass

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Question

$\sum \limits_{n=0}^{\infty} \frac{n^3((2 n) !)+(2 n-1)(n !)}{(n !)((2 n) !)}$

$=\sum \limits_{n=0}^{\infty} \frac{1}{(n-3) !}+\sum \limits_{n=0}^{

Vedclass · Accepted Answer

$26$

Vedclass · Answer

$\sum \limits_{n=0}^{\infty} \frac{n^3((2 n) !)+(2 n-1)(n !)}{(n !)((2 n) !)}$

$=\sum \limits_{n=0}^{\infty} \frac{1}{(n-3) !}+\sum \limits_{n=0}^{\infty} \frac{3}{(n-2) !}$

$+\sum \limits_{n=0}^{\infty} \frac{1}{(n-1) !}+\sum \limits_{n=0}^{\infty} \frac{1}{(2 n-1) !}-\sum \limits_{n=0}^{\infty} \frac{1}{(2 n) !}$

$=e+3 e+e+\frac{1}{2}\left(e-\frac{1}{e}\right)-\frac{1}{2}\left(e+\frac{1}{e}\right)$

$=5 e-\frac{1}{e}$

$a^2-b+c=26$

Let $\quad \sum \limits_{n=0}^{\infty} \frac{n^3((2 n) !)+(2 n-1)(n !)}{(n !)((2 n) !)}=a e+\frac{b}{e}+c$, where $a, b, c \in Z$ and $e=\sum \limits_{n=0}^{\infty} \frac{1}{n!}$ Then $a^2-b+c$ is equal to $................$.

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