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Question: \[1 + \frac{1 + 2}{2!} + \frac{1 + 2 + 3}{3!} + \frac{1 + 2 + 3 + 4}{4!} + ........\infty =\]...

1+1+22!+1+2+33!+1+2+3+44!+........∞=1 + \frac{1 + 2}{2!} + \frac{1 + 2 + 3}{3!} + \frac{1 + 2 + 3 + 4}{4!} + ........\infty =

A

e

B

3 e

C

e/2

D

3e/2

Answer

3e/2

Explanation

Solution

Tn=∑nn!=n(n+1)2.n!T_{n} = \frac{\sum_{}^{}n}{n!} = \frac{n(n + 1)}{2.n!}

=12[(n+1)(n−1)!]=12[n−1(n−1)!+2(n−1)!]= \frac{1}{2}\left\lbrack \frac{(n + 1)}{(n - 1)!} \right\rbrack = \frac{1}{2}\left\lbrack \frac{n - 1}{(n - 1)!} + \frac{2}{(n - 1)!} \right\rbrack

=12[1(n−2)!+2(n−1)!]= \frac{1}{2}\left\lbrack \frac{1}{(n - 2)!} + \frac{2}{(n - 1)!} \right\rbrack

Sn=∑n=1∞Tn=12∑n=1∞1(n−2)!+∑n=1∞1(n−1)!=e2+e=3e2Sn = \sum_{n = 1}^{\infty}{T_{n} = \frac{1}{2}\sum_{n = 1}^{\infty}{\frac{1}{(n - 2)!} + \sum_{n = 1}^{\infty}\frac{1}{(n - 1)!}}} = \frac{e}{2} + e = \frac{3e}{2}