Sequence And Series Question 181
Question: $ 1+\frac{1+2}{2,!}+\frac{1+2+3}{3,!}+\frac{1+2+3+4}{4,!}+….\infty = $
[Roorkee 1999; MP PET 2003]
Options:
A) $ e $
B) $ 3,e $
C) $ e/2 $
D) $ 3e/2 $
Show Answer
Answer:
Correct Answer: D
Solution:
$ T_{n}=\frac{\Sigma n}{n,!}=\frac{n(n+1)}{2,(n),!} $ $ =\frac{1}{2}[ \frac{(n+1)}{(n-1)!} ]=\frac{1}{2}[ \frac{n-1}{(n-1)!}+\frac{2}{(n-1)!} ] $ $ =\frac{1}{2}[ \frac{1}{(n-2)!}+\frac{2}{(n-1)!} ]=\frac{(e+2e)}{2}=\frac{3e}{2} $ .
BETA
