Sequence And Series Question 157
Question: $ 1+\frac{2^{3}}{2,!}+\frac{3^{3}}{3,!}+\frac{4^{3}}{4,!}+….\infty $ =
[MNR 1976; MP PET 1997]
Options:
A) $ 2,e $
B) $ 3,e $
C) $ 4,e $
D) $ 5,e $
Show Answer
Answer:
Correct Answer: D
Solution:
$ S=\frac{1^{3}}{1,!}+\frac{2^{3}}{2\ !}+\frac{3^{3}}{3\ !}+……..+\frac{n^{3}}{n\ !}+…… $ Here $ T_{n}=\frac{n^{3}}{n\ !}=\frac{n^{2}}{(n-1)\ !}=\frac{n^{2}-1}{(n-1)\ !}+\frac{1}{(n-1)\ !} $ $ =\frac{n+1}{(n-2)\ !}+\frac{1}{(n-1)\ !}=\frac{n}{(n-2)\ !}+\frac{1}{(n-2)\ !}+\frac{1}{(n-1)\ !} $ $ =\frac{n-2}{(n-2)\ !}+\frac{2}{(n-2)\ !}+\frac{1}{(n-2)\ !}+\frac{1}{(n-1)\ !} $ $ =\frac{1}{(n-3)\ !}+\frac{3}{(n-2)\ !}+\frac{1}{(n-1)\ !} $ Hence, sum = $ \sum\limits_{n=1}^{\infty }{\frac{1}{(n-3),!}}+3\sum\limits_{n=1}^{\infty }{\frac{1}{(n-2),!}}+\sum\limits_{n=1}^{\infty }{\frac{1}{(n-1),!}} $ = $ e+3e+e=5e $ .
BETA
