Sequence And Series Question 500
Question: $ 1+\frac{2}{3,!}+\frac{3}{5,!}+\frac{4}{7,!}+……\infty =, $
Options:
A) e
B) $ 2,e $
C) e/2
D) e/3
Show Answer
Answer:
Correct Answer: C
Solution:
$ S=\frac{1}{1\ !}+\frac{2}{3\ !}+\frac{3}{5\ !}+\frac{4}{7\ !}+…….+\frac{n}{(2n-1)\ !}+….. $ Here $ T_{n}=\frac{1}{2}.\frac{2n}{(2n-1)\ !}=\frac{1}{2}\frac{(2n-1)+1}{(2n-1)\ !} $ $ =\frac{1}{2}{ \frac{1}{(2n-2)\ !}+\frac{1}{(2n-1)\ !} } $
$ \Rightarrow S=\sum{T_{n}=\frac{1}{2}{ \frac{e+{e^{-1}}}{2}+\frac{e-{e^{-1}}}{2} }=\frac{e}{2}} $ . Trick: The sum of this series upto 4 terms is 1.359 …… and this is value of e/2 approximately.
BETA
