1+1+52!+1+5+523!+……∞=

1+1+52+1+5+523+……∞ =∑1∞ 1+5+52+...+5n-1n =∑1∞ 5n-14 n =14 ∑1∞ 5nn-1n =14 e5-1-(e-1)=e5-e4

1+1+52!+1+5+523!+……∞=

1+1+52+1+5+523+……∞ =∑1∞ 1+5+52+...+5n-1n =∑1∞ 5n-14 n =14 ∑1∞ 5nn-1n =14 e5-1-(e-1)=e5-e4

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

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$\frac{1}{3} e (e^{4} + 1)$

$\frac{1}{3} e (e^{4} - 1)$

$\frac{1}{4} e (e^{4} + 1)$

$\frac{1}{4} e (e^{4} - 1)$

answer is B.

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Detailed Solution

$\begin{array}{rcl} 1 + \frac{1 + 5}{2} + \frac{1 + 5 + 5^{2}}{3} + \dots \dots \infty \\ = & \sum_{1}^{\infty} \frac{1 + 5 + 5^{2} + . . . + 5^{n - 1}}{n} \\ = & \sum_{1}^{\infty} \frac{5^{n} - 1}{4 n} \\ = & \frac{1}{4} (\sum_{1}^{\infty} \frac{5^{n}}{n} - \frac{1}{n}) \\ = & \frac{1}{4} ((e^{5} - 1) - (e - 1)) = \frac{e^{5} - e}{4} \end{array}$