The sum of the series \sum_{n=1}^{\infty} \frac{n^{2}+6 n+10}{(2 n+1) !} is equal to [26 Feb 2021 Shift 2]

Correct Answer is : 418e−198e−1−10

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

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

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

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

JEE Main 26 Feb 2021 Shift 2 Solved Paper

Section: Mathematics

Question : 74 of 89

Marks: +1, -0

The sum of the series

∞

∑

n=1

‌

n2+6n+10

(2n+1)!

is equal to

[26 Feb 2021 Shift 2]

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