Sequence And Series Question 548
Question: The sum of the series $ 1+\frac{1}{4.2,!}+\frac{1}{16.4,!}+\frac{1}{64.6,!}+….. $ and inf. is
[AIEEE 2005]
Options:
A) $ \frac{e-1}{2\sqrt{e}} $
B) $ \frac{e+1}{2\sqrt{e}} $
C) $ \frac{e-1}{\sqrt{e}} $
D) $ \frac{e+1}{\sqrt{e}} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \frac{e^{x}+{e^{-x}}}{2}=1+\frac{x^{2}}{2,!}+\frac{x^{4}}{4,!}+\frac{x^{6}}{6,!}+….\infty $ Putting $ x=\frac{1}{2} $ , we get $ 1+\frac{1}{4,.,2,!}+\frac{1}{16.4,!}+\frac{1}{64.6!}+…\infty $ $ =\frac{{e^{1/2}}+{e^{-1/2}}}{2}=\frac{e+1}{2\sqrt{e}} $ .
BETA
