Binomial Theorem And Its Simple Applications Question 47
Question: $ \frac{1}{1.2}+\frac{1.3}{1.2.3.4}+\frac{1.3.5}{1.2.3.4.5.6}+….\infty = $
Options:
A) $ \sqrt{e} $
B) $ \sqrt{e}+1 $
C) $ \sqrt{e}-1 $
D) $ e-1 $
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] $ T_{n}=\frac{1.3.5…….(2n-1)}{(2n)!} $
$ =\frac{1.2.3.4….(2n-1).2n}{(2n)!2.,4.,6…..2n} $
$ =\frac{(2n)!}{(2n)!2^{n}n!}=\frac{1}{2^{n}n!} $ Now putting $ n=1,2,3,……… $ we see that the sum of series $ S=\frac{1}{2}+\frac{{{(1/2)}^{2}}}{2!}+\frac{{{(1/2)}^{3}}}{3!}+…. $
$ ={e^{\frac{1}{2}}}-1=\sqrt{e}-1 $
BETA
