The sum of the series sum_{n=1}^{infty} frac{ | vedclass

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(B) Let $T_n = \frac{n^2+6n+10}{(2n+1)!}$.We can rewrite the numerator as:$n^2+6n+10 = \frac{1}{4}(4n^2+24n+40) = \frac{1}{4}((2n+1)^2 + 20n + 39) = \

Vedclass · Accepted Answer

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

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(B) Let $T_n = \frac{n^2+6n+10}{(2n+1)!}$.We can rewrite the numerator as:$n^2+6n+10 = \frac{1}{4}(4n^2+24n+40) = \frac{1}{4}((2n+1)^2 + 20n + 39) = \frac{1}{4}((2n+1)^2 + 10(2n+1) + 29)$.Thus,$T_n = \frac{1}{4} \left[ \frac{2n+1}{(2n)!} + \frac{10}{(2n)!} + \frac{29}{(2n+1)!} \right]$.Using $\frac{2n+1}{(2n)!} = \frac{2n}{(2n)!} + \frac{1}{(2n)!} = \frac{1}{(2n-1)!} + \frac{1}{(2n)!}$,we get:$T_n = \frac{1}{4} \left[ \frac{1}{(2n-1)!} + \frac{11}{(2n)!} + \frac{29}{(2n+1)!} \right]$.Summing from $n=1$ to $\infty$:$S = \frac{1}{4} \left[ \sum_{n=1}^{\infty} \frac{1}{(2n-1)!} + 11 \sum_{n=1}^{\infty} \frac{1}{(2n)!} + 29 \sum_{n=1}^{\infty} \frac{1}{(2n+1)!} \right]$.Recall the series expansions: $\sum_{n=1}^{\infty} \frac{1}{(2n-1)!} = \frac{e-e^{-1}}{2}$,$\sum_{n=1}^{\infty} \frac{1}{(2n)!} = \frac{e+e^{-1}-2}{2}$,and $\sum_{n=1}^{\infty} \frac{1}{(2n+1)!} = \frac{e-e^{-1}-2}{2}$.Substituting these values:$S = \frac{1}{4} \left[ \frac{e-e^{-1}}{2} + 11 \frac{e+e^{-1}-2}{2} + 29 \frac{e-e^{-1}-2}{2} \right]$$S = \frac{1}{8} [e - e^{-1} + 11e + 11e^{-1} - 22 + 29e - 29e^{-1} - 58]$$S = \frac{1}{8} [41e - 19e^{-1} - 80] = \frac{41}{8}e - \frac{19}{8}e^{-1} - 10$.

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

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