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माना $\sum_{n=0}^{\infty} \frac{n^3((2 n) !)+(2 n-1)(n !)}{(n !)((2 n) !)}=a e+\frac{b}{e}+c\ $है, जहाँ $\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathbb{Z}$ तथा $\mathrm{e}=\sum_{\mathrm{n}=0}^{\infty} \frac{1}{\mathrm{n} !}$ है तो $\mathrm{a}^2-\mathrm{b}+\mathrm{c}$ बराबर है
A
$25$
B
$24$
C
$23$
D
$26$
(JEE MAIN-2023)
Solution
$\sum \limits_{n=0}^{\infty} \frac{n^3((2 n) !)+(2 n-1)(n !)}{(n !)((2 n) !)}$
$=\sum \limits_{n=0}^{\infty} \frac{1}{(n-3) !}+\sum \limits_{n=0}^{\infty} \frac{3}{(n-2) !}$
$+\sum \limits_{n=0}^{\infty} \frac{1}{(n-1) !}+\sum \limits_{n=0}^{\infty} \frac{1}{(2 n-1) !}-\sum \limits_{n=0}^{\infty} \frac{1}{(2 n) !}$
$=e+3 e+e+\frac{1}{2}\left(e-\frac{1}{e}\right)-\frac{1}{2}\left(e+\frac{1}{e}\right)$
$=5 e-\frac{1}{e}$
$a^2-b+c=26$
Standard 11
Mathematics
