Sequence And Series Question 340
Question: $ \frac{1+\frac{2^{2}}{2,!}+\frac{2^{4}}{3,!}+\frac{2^{6}}{4,!}+…..\infty }{1+\frac{1}{2,!}+\frac{2}{3,!}+\frac{2^{2}}{4,!}+….\infty }= $
Options:
A) $ e^{2} $
B) $ e^{2}-1 $
C) $ {e^{3/2}} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \frac{1+\frac{2^{2}}{2\ !}+\frac{2^{4}}{3\ !}+\frac{2^{6}}{4\ !}+….\infty }{1+\frac{1}{2\ !}+\frac{2}{3\ !}+\frac{2^{2}}{4\ !}+…..\infty } $ $ =\frac{\frac{1}{2^{2}}{ \frac{2^{2}}{1\ !}+\frac{{{(2^{2})}^{2}}}{2\ !}+\frac{{{(3^{2})}^{3}}}{3\ !}+…… }}{\frac{1}{2^{2}}{ 2+2+\frac{2^{2}}{2\ !}+\frac{2^{3}}{3\ !}+…… }}=\frac{{e^{(2^{2})}}-1}{1+e^{2}}=e^{2}-1 $ .
BETA
