Sequence And Series Question 549
Question: The sum of the series $ \frac{1^{2}}{1\cdot 2,!}+\frac{1^{2}+2^{2}}{2\cdot 3,!}+\frac{1^{2}+2^{2}+3^{2}}{3\cdot 4,!}+..+\frac{1^{2}+2^{2}+…+n^{2}}{n\cdot (n+1),!}+…\infty $ equals
[AMU 2002]
Options:
A) $ e^{2} $
B) $ \frac{1}{2}{{(e+{e^{-1}})}^{2}} $
C) $ \frac{3e-1}{6} $
D) $ \frac{4e+1}{6} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ T_{n}=\frac{n(n+1)(2n+1)}{6n(n+1)!} $
$ \therefore S=\frac{1}{6}\sum{[ \frac{2n+1}{(n)!} ]} $ = $ \frac{1}{6}\sum{[ 2.\frac{1}{(n-1)!}+\frac{1}{(n)!} ]} $ = $ \frac{1}{6}[2.e+e-1]=\frac{1}{6}[3e-1] $ .
BETA
