If an=1+2+3+……+n2! then a1+a2+a3+……∞=

a1+a2+a3+……∞ =∑1∞ an =∑1∞ (1+2+3+...+n)n =∑1∞ n(n+1)2 n=∑1∞ n+12 n-1 =12 ∑1∞ (n-1)+2n-1 =12 ∑1∞ 1n-2+2n-1 =12 (e+2e)=3e2

If an=1+2+3+……+n2! then a1+a2+a3+……∞=

a1+a2+a3+……∞ =∑1∞ an =∑1∞ (1+2+3+...+n)n =∑1∞ n(n+1)2 n=∑1∞ n+12 n-1 =12 ∑1∞ (n-1)+2n-1 =12 ∑1∞ 1n-2+2n-1 =12 (e+2e)=3e2

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If $a_{n} = \frac{1 + 2 + 3 + \dots \dots + n}{2!}$ then $a_{1} + a_{2} + a_{3} + \dots \dots \infty =$

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$\frac{7 e}{2}$

$\frac{3 e}{2}$

$\frac{5 e}{2}$

$\frac{e}{2}$

answer is B.

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Detailed Solution

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