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Question: \[\frac{2}{1!} + \frac{4}{3!} + \frac{6}{5!} + \frac{8}{7!} + .........\infty =\]...

21!+43!+65!+87!+.........∞=\frac{2}{1!} + \frac{4}{3!} + \frac{6}{5!} + \frac{8}{7!} + .........\infty =

A

1/e1/e

B

e

C

2e

D

3e

Answer

e

Explanation

Solution

21!+43!+65!+87!+.........∞=(1+1)1!+(1+3)3!+(1+5)5!+(1+7)7!+........∞=(11!+13!+15!+17!+........∞)+(1+12!+14!+16!+........∞)=e−e−12+e+e−12=e\frac{2}{1!} + \frac{4}{3!} + \frac{6}{5!} + \frac{8}{7!} + .........\infty = \frac{(1 + 1)}{1!} + \frac{(1 + 3)}{3!} + \frac{(1 + 5)}{5!} + \frac{(1 + 7)}{7!} + ........\infty = \left( \frac{1}{1!} + \frac{1}{3!} + \frac{1}{5!} + \frac{1}{7!} + ........\infty \right) + \left( 1 + \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + ........\infty \right) = \frac{e - e^{- 1}}{2} + \frac{e + e^{- 1}}{2} = e