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

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(C) We have the sum $S = \sum_{n=1}^{\infty} \frac{2n^2+n+1}{n!}$.Since $n^2 = n(n-1) + n$,we can write the numerator as $2n(n-1) + 2n + n + 1 = 2n(n-

Vedclass · Accepted Answer

$6e-1$

Vedclass · Answer

(C) We have the sum $S = \sum_{n=1}^{\infty} \frac{2n^2+n+1}{n!}$.Since $n^2 = n(n-1) + n$,we can write the numerator as $2n(n-1) + 2n + n + 1 = 2n(n-1) + 3n + 1$.Thus,$\frac{2n^2+n+1}{n!} = \frac{2n(n-1)}{n!} + \frac{3n}{n!} + \frac{1}{n!} = \frac{2}{(n-2)!} + \frac{3}{(n-1)!} + \frac{1}{n!}$.Summing from $n=1$ to $\infty$:$S = \sum_{n=2}^{\infty} \frac{2}{(n-2)!} + \sum_{n=1}^{\infty} \frac{3}{(n-1)!} + \sum_{n=1}^{\infty} \frac{1}{n!}$.Using the series expansion $e = \sum_{k=0}^{\infty} \frac{1}{k!}$:$S = 2e + 3e + (e-1) = 6e - 1$.

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

Similar Questions

Let $\sum_{n=0}^{\infty} \frac{n^3((2n)!) + (2n-1)(n!)}{(n!)((2n)!)} = ae + \frac{b}{e} + c$,where $a, b, c \in \mathbb{Z}$ and $e = \sum_{n=0}^{\infty} \frac{1}{n!}$. Then $a^2 - b + c$ is equal to $................$.

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