Question

If $\frac{e^{5x}+e^x}{e^{3x}}$ is expanded in a series of ascending powers of $x$ and $n$ is an odd natural number, then the coefficient of $x^n$ is

• A. $\frac{2^n}{n!}$
• B. $\frac{2^{n+1}}{2n!}$
• C. $\frac{2^{2n}}{\left(2n\right)!}$
• D. $\frac{2^{2n-1}}{(2n-1)!}$
• E. None of these

Answer 1

Answer: (E)

None of these

$\frac{e^{5x}+e^x}{e^{3x}}$
$=e^{2x}+e^{-2x}$
$=(1+2x+\frac{(2x{)}^2}{2!}+\frac{(2x{)}^3}{3!}+...\infty )+(1-2x+\frac{(2x{)}^2}{2!}-\frac{(2x{)}^3}{3!}+...\infty )$
$=2[1+\frac{(2x{)}^2}{2!}+\frac{(2x{)}^4}{4!}...\infty ]$
Hence there won't be any term with odd power of $x$