Sequence And Series Question 566
Question: $ 1+\frac{1+3}{2,!}+\frac{1+3+5}{3,!}+\frac{1+3+5+7}{4,!}+…….\infty = $
Options:
A) $ e/2 $
B) $ e $
C) $ 2,e $
D) $ 3e $
Show Answer
Answer:
Correct Answer: C
Solution:
$ T_{n}=\frac{1+3+5+…….+(2n-1)}{n\ !}=\frac{\frac{n}{2}[2\ .1+(n-1)2]}{n\ !} $ $ =\frac{n-1+1}{(n-1)\ !}=\frac{1}{(n-2)\ !}+\frac{1}{(n-1)\ !} $
$ \Rightarrow S=\sum\limits_{n=1}^{\infty }{T_{n}}=e+e=2e $ .
BETA
