Sequence And Series Question 535
Question: $ \frac{\frac{1}{2,!}+\frac{1}{4,!}+\frac{1}{6,!}+…..\infty }{1+\frac{1}{3,!}+\frac{1}{5,!}+\frac{1}{7,!}+…..\infty }= $
Options:
A) $ \frac{e+1}{e-1} $
B) $ \frac{e-1}{e+1} $
C) $ \frac{e^{2}+1}{e^{2}-1} $
D) $ \frac{e^{2}-1}{e^{2}+1} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \frac{\frac{1}{2\ !}+\frac{1}{4\ !}+\frac{1}{6\ !}+……\infty }{1+\frac{1\ }{3\ !}+\frac{1}{5\ !}+\frac{1}{7\ !}+…….\infty } $ $ =,\frac{2[ \frac{1}{2,!}+\frac{1}{4,!}+\frac{1}{6,!}+….\infty ]}{2,[ 1+\frac{1}{3,!}+\frac{1}{5,!}+\frac{1}{7,!}+….\infty ]} $ $ =\frac{(e+{e^{-1}})-2}{(e-{e^{-1}})}=\frac{e+\frac{1}{e}-2}{e-\frac{1}{e}}=\frac{e^{2}+1-2e}{e^{2}-1} $ $ =\frac{{{(e-1)}^{2}}}{(e-1)(e+1)}=\frac{e-1}{e+1} $ .
BETA
