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

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Answer: B. {{41} \over 8}e - {{19} \over 8}{e^{ - 1}} - 10 \sum\limits_{n = 1}^\infty {{{{n^2} + 6n + 10} \over {(2n + 1)!}}} Put 2n + 1 = r, where r = 3, 5, 7, ....... \Rightarrow n = {{r - 1} \over 2} {{{n^2} + 6n + 10} \over {(2n + 1)!}} = {{{{\left( {{{r - 1} \over 2}} \right)}^2} + 3r - 3 + 10} \over {r!}} = {{{r^2} + 10r + 29} \over {4r!}} = {{{r(r - 1) + 11r + 29} \over {4r!}}} Now, \sum\limits_{r = 3,5,7} {{{r(r - 1) + 11r + 29} \over {4r!}}} ={1 \over 4}\sum\limits_{r = 3,5,7,......} {\left( {{1 \over {(r - 2)!}} + {{11} \over {(r - 1)!}} + {{29} \over {r!}}} \right)} = {1 \over 4}\left\{ {\left( {{1 \over {1!}} + {1 \over {3!}} + {1 \over {5!}} + ......} \right) + 11\left( {{1 \over {2!}} + {1 \over {4!}} + {1 \over {6!}} + ......} \right) + 29\left( {{1 \over {3!}} + {1 \over {5!}} + {1 \over {7!}} + ......} \right)} \right\} = {1 \over 4}\left\{ {{{e - {1 \over e}} \over 2} + 11\left( {{{e + {1 \over e} - 2} \over 2}} \right) + 29\left( {{{e - {1 \over e} - 2} \over 2}} \right)} \right\} = {1 \over 8}\left\{ {e - {1 \over e} + 11e + {{11} \over e} - 22 + 29e - {{29} \over e} - 58} \right\} = {1 \over 8}\left\{ {41e - {{19} \over e} - 80} \right\}

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

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