sum_{n=1}^{infty} frac{2n^2+n+1}{n!} is equal | vedclass

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(C) Let $S = \sum_{n=1}^{\infty} \frac{2n^2+n+1}{n!}$.We can write $2n^2+n+1$ in terms of $n(n-1)$,$n$,and constants:$2n^2+n+1 = 2n(n-1) + 3n + 1$.Thu

Vedclass · Accepted Answer

$6e-1$

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(C) Let $S = \sum_{n=1}^{\infty} \frac{2n^2+n+1}{n!}$.We can write $2n^2+n+1$ in terms of $n(n-1)$,$n$,and constants:$2n^2+n+1 = 2n(n-1) + 3n + 1$.Thus,the sum becomes:$S = \sum_{n=1}^{\infty} \frac{2n(n-1) + 3n + 1}{n!} = \sum_{n=1}^{\infty} \frac{2n(n-1)}{n!} + \sum_{n=1}^{\infty} \frac{3n}{n!} + \sum_{n=1}^{\infty} \frac{1}{n!}$.For $n=1$,the first term is $0$. For $n=1, 2$,the second term is $3/1! = 3$. For $n=1$,the third term is $1/1! = 1$.$S = \sum_{n=2}^{\infty} \frac{2}{(n-2)!} + \sum_{n=1}^{\infty} \frac{3}{(n-1)!} + \sum_{n=1}^{\infty} \frac{1}{n!}$.Using the series $e = \sum_{k=0}^{\infty} \frac{1}{k!}$:$S = 2e + 3e + (e-1) = 6e-1$.

$\sum_{n=1}^{\infty} \frac{2n^2+n+1}{n!}$ is equal to

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