Mathematics · 2021
JEE Main · 26 February 2021, Shift 2 · Q66
The sum of the series Σ_n=1^∞ (n^2+6 n+10)/((2 n+1)!) is equal to:
The sum of the series $\displaystyle \sum_{n=1}^{\infty} \frac{n^2+6 n+10}{(2 n+1)!}$ is equal to :
Official answer
(4)
$\displaystyle \frac{41}{8} e-\frac{19}{8} e^{-1}-10$
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JEE Main 2021 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.