Sequence And Series Question 463
Question: $ 1+\frac{4^{2}}{3,!}+\frac{4^{4}}{5,!}+……\infty = $
Options:
A) $ \frac{e^{4}+{e^{-4}}}{4} $
B) $ \frac{e^{4}-{e^{-4}}}{4} $
C) $ \frac{e^{4}+{e^{-4}}}{8} $
D) $ \frac{e^{4}-{e^{-4}}}{8} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ S=1+\frac{4^{2}}{3\ !}+\frac{4^{4}}{5\ !}+……..\infty $ $ =\frac{1}{4}{ 4+\frac{4^{3}}{3\ !}+\frac{4^{5}}{5,!}+…… }=\frac{1}{4}( \frac{e^{4}-{e^{-4}}}{2} ) $ .
BETA
