Sequence And Series Question 89
Question: $ \frac{1^{2}.2}{1,!}+\frac{2^{2}.3}{2,!}+\frac{3^{2}.4}{3,!}+…..\infty = $
[UPSEAT 1999]
Options:
A) $ 6,e $
B) $ 7,e $
C) $ 8,e $
D) $ 9,e $
Show Answer
Answer:
Correct Answer: B
Solution:
$ S=\frac{1^{2}.2}{1!}+\frac{2^{2}.3}{2!}+\frac{3^{2}.4}{3!}+…. $ Here $ T_{n}=\frac{n^{2}.(n+1)}{n!}=\frac{n(n+1)}{(n-1)!} $ $ =\frac{(n-1)(n-2)+4n-2}{(n-1)!}=\frac{1}{(n-3)!}+\frac{4(n-1)+2}{(n-1)!} $ $ \frac{1}{(n-3)!} $
$ \Rightarrow ,S=\sum{T_{n}=e+4e+2e=7e} $ .
BETA
