Binomial-Theorem-And-Its-Simple-Applications Question 219
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
Correct Answer: D $ \therefore ,AB=e\cdot e^{-1}=1 $
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Answer:
Solution:
$ =\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.
BETA
