∑n=1∞ 2n2+n+1n =∑n=1∞ 2nn-1+1n-1+1n =∑n=1∞ (n-1)+1n-1+e+(e-1) =2∑n=1∞ 1n-2+1n-1+2e-1 =2e+e+2e-1 =6e-1

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

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2e – 1

2e + 1

6e – 1

6e + 1

answer is C.

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Detailed Solution

$\begin{array}{rcl} \sum_{n = 1}^{\infty} \frac{2 n^{2} + n + 1}{n} \\ = & \sum_{n = 1}^{\infty} (\frac{2 n}{n - 1} + \frac{1}{n - 1} + \frac{1}{n}) \\ = & \sum_{n = 1}^{\infty} \frac{(n - 1) + 1}{n - 1} + e + (e - 1) \\ = & 2 [\sum_{n = 1}^{\infty} (\frac{1}{n - 2} + \frac{1}{n - 1})] + 2 e - 1 \\ = & 2 [(e + e)] + 2 e - 1 \\ = & 6 e - 1 \end{array}$