Sequence And Series Question 278
Question: The sum of $ \frac{2}{1,!}+\frac{6}{2,!}+\frac{12}{3,!}+\frac{20}{4,!}+ $ …….is
[UPSEAT 2000]
Options:
A) $ \frac{3e}{2} $
B) $ e $
C) $ 2e $
D) $ 3e $
Show Answer
Answer:
Correct Answer: D
Solution:
Let $ S=\frac{2}{1!}+\frac{6}{2!}+\frac{12}{3!}+\frac{20}{4!}+….. $ and let $ S_1=2+6+12+20+…..+T_{n} $ $ S_1=,2+6+12+……………..{T_{n-1}}+T_{n} $ $ \underline{\overline{0=2+4+6+8+……uptonterms-T_{n}}} $ $ T_{n}=2+4+6+8+…….\text{upto }nterms $
Þ $ T_{n} $ = $ \frac{n}{2}[2\times 2+(n-1),2] $ = $ n(2+n-1)=n(n+1) $ \ nth term of given series $ T_{n}=\frac{n(n+1)}{n!}\text{or }T_{n}=\frac{n(n+1)}{n(n-1)!} $ or $ T_{n}=\frac{1}{(n-2),!}+\frac{2}{(n-1)!} $ Now, sum = $ \sum\limits_{n=1}^{\infty }{\frac{1}{(n-2)!}+2\sum\limits_{n=1}^{\infty }{\frac{1}{(n-1),!}}}=e+2e=3e $ .
BETA
