Sequence And Series Question 77
Question: $ ( 1+\frac{1}{2,!}+\frac{1}{4,!}+…. )( 1+\frac{1}{3,!}+\frac{1}{5,!}+…. ),= $
Options:
A) $ e^{4} $
B) $ \frac{e^{2}-1}{e^{2}} $
C) $ \frac{e^{4}-1}{4,e^{2}} $
D) $ \frac{e^{4}+1}{4,e^{2}} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ ( 1+\frac{1}{2\ !}+\frac{1}{4\ !}+….. ),( 1+\frac{1}{3\ !}+\frac{1}{5\ !}+……. ) $ $ =\frac{e+{e^{-1}}}{2}.\frac{e-{e^{-1}}}{2}=\frac{e^{2}-{e^{-2}}}{4}=\frac{e^{4}-1}{4e^{2}} $ .
BETA
