Sequence And Series Question 110
Question: The sum of the series $ 1+\frac{1}{4.2!}+\frac{1}{16.4!}+\frac{1}{64.6!}+… $ is
Options:
A) $ \frac{e-1}{\sqrt{e}} $
B) $ \frac{e+1}{\sqrt{e}} $
C) $ \frac{e-1}{2\sqrt{e}} $
D) $ \frac{e+1}{2\sqrt{e}} $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] We know that $ e^{x}=1+\frac{x}{1!}+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+…\infty $ ….(i)
$ \therefore {e^{-x}}=1-\frac{x}{1!}+\frac{x^{2}}{2!}-\frac{x^{3}}{3!}+\frac{x^{4}}{4}-…\infty $ …(ii) On adding equations (i) and (ii), we get
$ \therefore e^{x}+{e^{-x}}=2x+\frac{2x^{2}}{2!}+\frac{2x^{4}}{4!}+…\infty $
$ \therefore \frac{e^{x}+{e^{-x}}}{2}=1+\frac{x^{2}}{2!}+\frac{x^{4}}{4}+\frac{x^{6}}{6!}+… $ Putting x=1/2, we get $ \frac{e+1}{2\sqrt{e}}=1+\frac{1}{4.2!}+\frac{1}{16.4!}+\frac{1}{16.4!}+\frac{1}{64.6!}+ $
BETA
