1+1+22!+1+2+223!+……∞=

1+1+22+1+2+223+……∞ =∑1∞ 1+2+22+...+2n-1n =∑1∞ 12n-1n =∑1∞2nn-1n =e2-1-(e-1) =e2-e=e(e-1)

1+1+22!+1+2+223!+……∞=

1+1+22+1+2+223+……∞ =∑1∞ 1+2+22+...+2n-1n =∑1∞ 12n-1n =∑1∞2nn-1n =e2-1-(e-1) =e2-e=e(e-1)

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

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e(e –1)

$\frac{e (e - 1)}{2}$

$\frac{e (e^{2} - 1)}{2}$

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

answer is A.

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Detailed Solution

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