Question

The coefficient of $x^n$ in the expansion of ${(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+....\frac{x^n}{n!})}^2$ equals

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

Answer 1

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

${(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+....\frac{x^n}{n!})}^2=(e^x{)}^2=e^{2x}$
$={(1+\frac{2x}{1!}+\frac{(2x{)}^2}{2!}+\frac{(2x{)}^3}{3!}+....\frac{(2x{)}^n}{n!})}^2$
Hence coefficient of $x^n$ is $\frac{2^n}{n!}$