Sequence And Series Question 377
Question: $ \frac{2}{1,!}+\frac{4}{3,!}+\frac{6}{5,!}+\frac{8}{7,!}+……\infty = $
[JMI CET 2000]
Options:
A) $ 1/e $
B) $ e $
C) $ 2,e $
D) $ 3e $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \frac{2}{1!}+\frac{4}{3!}+\frac{6}{5!}+\frac{8}{7!}+….\infty $ $ =\frac{(1+1)}{1!}+\frac{(1+3)}{3!}+\frac{(5+1)}{5!}+\frac{(7+1)}{7!}+….\infty $ $ =( \frac{1}{1!}+\frac{1}{3!}+\frac{1}{5!}+\frac{1}{7!}+….\infty )+( 1+\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+….\infty ) $ $ =\frac{e-{e^{-1}}}{2}+\frac{e+{e^{-1}}}{2}=e $ .
BETA
