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

• A. $\frac{2^n}{n!}$
• B. $\frac{2^n-2}{n!}$
• C. $\frac{2^{n-1}-1}{n!}$
• D. $\frac{2^{n-1}}{n!}$

Answer 1

The correct option is D $\frac{2^{n-1}}{n!}$

${(1+\frac{x^2}{2!}+\frac{x^4}{4!}+....)}^2={\left(\frac{e^x+e^{-x}}{2}\right)}^2$
$=\frac{1}{4}(e^{2x}+e^{-2x}+2)$
$=\frac{1}{4}\{2(1+\frac{(2x{)}^2}{2!}+\frac{(2x{)}^4}{4!}+.....)+2\}$
So, coefficient of $x^n$ (n even)$=\frac{1}{2}\left\{\frac{2^n}{n!}\right\}=\frac{2^{n-1}}{n!}$