The sum $\sum \limits_{n=1}^{\infty} \frac{2 n^2+3 n+4}{(2 n) !}$ is equal to :
The sum $\sum \limits_{n=1}^{\infty} \frac{2 n^2+3 n+4}{(2 n) !}$ is equal to :
- [JEE MAIN 2023]
- A
$\frac{11 e }{2}+\frac{7}{2 e }$
- B
$\frac{13 e }{4}+\frac{5}{4 e }-4$
- C
$\frac{11 e }{2}+\frac{7}{2 e }-4$
- D
$\frac{13 e }{4}+\frac{5}{4 e }$
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