Sequence And Series Question 546
Question: The sum of the series $ \frac{4}{1,!}+\frac{11}{2,!}+\frac{22}{3,!}+\frac{37}{4,!}+\frac{56}{5,!}+… $ is
[Kurukshetra CEE 2002]
Options:
A) 6 e
B) 6 e ? 1
C) 5 e
D) 5 e + 1
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ S=4+11+22+37+….+{T_{n-1}}+T_{n} $ or S = 4 + 11 + 22 + 37+??..+ $ {T_{n-1}}+T_{n} $
$ \therefore $ On subtracting we get $ 0=4+[7+11+15+19+….+(T_{n}-{T_{n-1}})]-T_{n} $ $ 0=4+\frac{n-1}{2}[14+(n-2)4]-T_{n} $
$ \therefore T_{n}=2n^{2}+n+1 $ Thus $ n{{}^{th}} $ term of given series is $ T_{n}=\frac{2n^{2}+n+1}{(n)!}=\frac{2n}{(n-1)!}+\frac{1}{(n-1)!}+\frac{1}{n!} $ $ =\frac{2(n-1+1)}{(n-1)!}+\frac{1}{(n-1)!}+\frac{1}{n!} $ $ =\frac{2}{(n-2)!}+\frac{3}{(n-1)!}+\frac{1}{n!} $
$ \therefore $ Sum $ =\sum\limits_{n=1}^{\infty }{T_{n}=2e+3e+e-1=6e-1} $ .
BETA
