Sequence And Series Question 401
Question: $ \frac{2}{3,!}+\frac{4}{5,!}+\frac{6}{7,!}+……\infty = $
[MNR 1979; MP PET 1995, 2002; Pb. CET 2002]
Options:
A) $ e $
B) $ 2,e $
C) $ e^{2} $
D) $ 1/e $
Show Answer
Answer:
Correct Answer: D
Solution:
$ S=\frac{2}{3\ !}+\frac{4}{5\ !}+\frac{6}{7\ !}+……+\frac{2n}{(2n+1)\ !}+…… $ Here $ T_{n}=\frac{(2n+1)-1}{(2n+1)\ !}=\frac{1}{(2n)\ !}-\frac{1}{(2n+1)\ !} $
$ \Rightarrow S=\sum\limits_{n=1}^{\infty }{T_{n}}=( \frac{1}{2\ !}+\frac{1}{4\ !}+\frac{1}{6\ !}+… ) $ $ -( \frac{1}{3\ !}+\frac{1}{5\ !}+\frac{1}{7\ !}+….. ) $ $ =( \frac{e+{e^{-1}}}{2}-1 )-( \frac{e-{e^{-1}}}{2}-1 )={e^{-1}}=\frac{1}{e} $ .
BETA
