Sequence And Series Question 584
Question: $ \frac{2}{1,!}+\frac{2+4}{2,!}+\frac{2+4+6}{3,!}+….\infty = $
[MNR 1985]
Options:
A) $ e $
B) $ 2,e $
C) $ 3,e $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ S=\frac{2}{1\ !}+\frac{2+4}{2\ !}+\frac{2+4+6}{3\ !}+……+\frac{\frac{n}{2}(2+2n)}{n\ !}+…..\infty $ Here $ T_{n}=\frac{n(n+1)}{n\ !}=\frac{n-1+2}{(n-1)\ !}=\frac{1}{(n-2)\ !}+\frac{2}{(n-1)\ !} $
$ \Rightarrow S=\sum\limits_{n=1}^{\infty }{T_{n}}=\sum\limits_{n=1}^{\infty }{\frac{1}{(n-2)\ !}+2\sum\limits_{n=1}^{\infty }{\frac{1}{(n-1)\ !}}} $ $ =e+2e=3e $ .
BETA
