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Question: \[\frac{2}{1!} + \frac{2 + 4}{2!} + \frac{2 + 4 + 6}{3!} + ....\infty =\]...

21!+2+42!+2+4+63!+....∞=\frac{2}{1!} + \frac{2 + 4}{2!} + \frac{2 + 4 + 6}{3!} + ....\infty =

A

ee

B

2e2e

C

3e3e

D

None of these

Answer

3e3e

Explanation

Solution

e4−e−48\frac{e^{4} - e^{- 4}}{8}

Here 1+221!+322!+423!+......∞=1 + \frac{2^{2}}{1!} + \frac{3^{2}}{2!} + \frac{4^{2}}{3!} + ......\infty =

2e2e 3e3e.