Sequence And Series Question 539
Question: If $ n $ is even, then in the expansion of $ {{( 1+\frac{x^{2}}{2,!}+\frac{x^{4}}{4,!}+…… )}^{2}} $ , the coefficient of $ x^{n} $ is
Options:
A) $ \frac{2^{n}}{n,!} $
B) $ \frac{2^{n}-2}{n!} $
C) $ \frac{{2^{n-1}}-1}{n,!} $
D) $ \frac{{2^{n-1}}}{n,!} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ {{( 1+\frac{x^{2}}{2!}+\frac{x^{4}}{4!}+…. )}^{2}}={{( \frac{e^{x}+{e^{-x}}}{2} )}^{2}} $ $ =\frac{1}{4}(e^{2x}+{e^{-2x}}+2) $ $ =\frac{1}{4}{ 2,( 1+\frac{{{(2x)}^{2}}}{2!}+\frac{{{(2x)}^{4}}}{4!}+…. )+2 } $ \ The coefficient of $ x^{n} $ (n even) $ =\frac{1}{2}{ \frac{2^{n}}{n!} }=\frac{{2^{n-1}}}{n!} $ .
BETA
