Sequence And Series Question 540
Question: $ 1+\frac{1+2}{1,!}+\frac{1+2+3}{2,!}+\frac{1+2+3+4}{3,!}+….\infty = $
Options:
0
1
C) $ \frac{7e}{2} $
D) $ 2,e $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \frac{1}{0!}+\frac{1+2}{1!}+\frac{1+2+3}{2!}+….\infty $ $ n^{th} $ term $ T_{n}=\frac{1+2+3+4+…..+n}{(n-1)!}=\frac{n(n+1)}{2(n-1)!} $ $ T_{n}=\frac{1}{2}[ \frac{1}{(n-1)!}+\frac{4}{(n-1)!}+\frac{2}{(n-1)!} ] $ Therefore sum $ {S_{\infty }}=\frac{7e}{2} $ .
BETA
