1+31!+52!+73!+……∞=

1+31+52+73+…… =1+∑1∞ (2n+1)n =1+∑1∞ (2)n-1+1n =1+∑1∞ 2n-1+1n =1+2e+(e-1) =3e

1+31!+52!+73!+……∞=

1+31+52+73+…… =1+∑1∞ (2n+1)n =1+∑1∞ (2)n-1+1n =1+∑1∞ 2n-1+1n =1+2e+(e-1) =3e

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$1 + \frac{3}{1!} + \frac{5}{2!} + \frac{7}{3!} + \dots \dots \infty =$

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answer is C.

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Detailed Solution

$\begin{array}{rcl} 1 + \frac{3}{1} + \frac{5}{2} + \frac{7}{3} + \dots \dots \\ = & {1 + \sum}_{1}^{\infty} \frac{(2 n + 1)}{n} \\ = & 1 + (\sum_{1}^{\infty} \frac{(2)}{n - 1} + \frac{1}{n}) \\ = & 1 + (\sum_{1}^{\infty} \frac{2}{n - 1} + \frac{1}{n}) \\ = & 1 + 2 e + (e - 1) \\ = & 3 e \end{array}$