Sequence-And-Series Question 669
Question: The sum of the infinite series $ \frac{2^{2}}{2!}+\frac{2^{4}}{4!}+\frac{2^{6}}{6!}+…. $ is equal to
Options:
A) $ \frac{e^{2}+1}{2e} $
B) $ \frac{e^{4}+1}{2e^{2}} $
C) $ \frac{{{(e^{2}-1)}^{2}}}{2e^{2}} $
D) $ \frac{{{(e^{2}+1)}^{2}}}{2e^{2}} $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] We know that $ \frac{e^{x}+{e^{-x}}}{2}=1+\frac{x^{2}}{2!}+\frac{x^{4}}{4!}+\frac{x^{6}}{6!}+…. $ keeping $ x=2, $ we get Expression $ =\frac{1}{2}[ \frac{e^{2}+{e^{-2}}}{2} ]-1=\frac{{{(e^{2}-1)}^{2}}}{2e^{2}} $
BETA
