Sequence And Series Question 596
Question: $ {{[ 1+\frac{1}{2,!}+\frac{1}{4,!}+….\infty ]}^{2}}-{{[ 1+\frac{1}{3,!}+\frac{1}{5,!}+…..\infty ]}^{2}}= $
Options:
A) 0
B) 1
C) $ -1 $
D) 2
Show Answer
Answer:
Correct Answer: B
Solution:
$ {{( \frac{e+{e^{-1}}}{2} )}^{2}}-{{( \frac{e-{e^{-1}}}{2} )}^{2}}=\frac{1}{4}{4,.,e,.,{e^{-1}}}=1 $ .
BETA
