Sequence And Series Question 476
Question: $ 1+\frac{2^{2}}{1,!}+\frac{3^{2}}{2,!}+\frac{4^{2}}{3,!}+……\infty = $
Options:
A) $ 2,e $
B) $ 3,e $
C) $ (0.5)-\frac{{{(0.5)}^{2}}}{2}+\frac{{{(0.5)}^{3}}}{3}-\frac{{{(0.5)}^{4}}}{4}+…. $
D) $ 5,e $
Show Answer
Answer:
Correct Answer: D
Solution:
$ S=1+\frac{2^{2}}{1\ !}+\frac{3^{2}}{2\ !}+\frac{4^{2}}{3\ !}+…….+\frac{n^{2}}{(n-1)\ !}+…….\infty $ Here $ T_{n}=\frac{n^{2}}{(n-1)\ !}=\frac{(n-1)(n-2)}{(n-1)\ !}+\frac{3(n-1)+1}{(n-1)\ !} $ $ =\frac{1}{(n-3)\ !}+\frac{3}{(n-2)\ !}+\frac{1}{(n-1)\ !} $
$ \Rightarrow S=\Sigma T_{n}=e+3e+e=5e $ .
BETA
