JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Analyzing +4 | -1

The sum of the series

\sum\limits_{n = 1}^\infty {\frac{{{n^2} + 6n + 10}}{{\left( {2n + 1} \right)!}}}

is equal to

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because the series can be decomposed and expressed in terms of exponential series expansions.

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