JEE Main 2021 (26 Feb, Shift 2) Maths Q66: The sum of the…

Question

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

Accepted Answer

Option 4: $\frac{41}{8}e - \frac{19}{8}e^{-1} - 10$

Answer

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

Answer

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

Answer

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

Answer

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

JEE Main 26 February 2021, Shift 2, Maths Q66