Binomial Theorem And Its Simple Applications Question 56
If $ A=\sum\limits_{n=1}^{\infty }{\frac{2n}{(2n-1)!}},\ B=\sum\limits_{n=1}^{\infty }{\frac{2n}{(2n+1)!}} $ then AB is equal to
Options:
A) $ e^{2} $
B) e
C) $ e + e^{2} $
D) 1
Show Answer
Answer:
Correct Answer: D
Solution:
- [d] $ A=\sum\limits_{n=1}^{\infty }{\frac{2n-1+1}{(2n-1)!}} $
$ =\sum\limits_{n=1}^{\infty }{[ \frac{1}{(2n-2)!}+\frac{1}{(2n-1)!} ]} $
$ =1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+….=e $
Similarly $ B={e^{-1}} $ as terms will be alternately positive and negative.
$ \therefore ,AB=e\cdot e^{-1}=1 $
BETA
