Question

In the expailsion of $\frac{e^{4x}-1}{e^{2x}}$ the coefficient of $x^2$ is:

• A. $\frac{1}{2}$
• B. 1
• C. 0
• D. none of these

Answer 1

Answer: (C)

$\frac{e^{4x}-1}{e^{2x}}=e^{2x}-e^{-2x}$
$=\left(1+\frac{2x}{1!}+\frac{(2x{)}^2}{2!}+\dots \right)$
$-\left(1-\frac{2x}{1!}+\frac{(2x{)}^2}{2!}-\frac{(2x{)}^3}{3!}+\dots \right)$
$=2\left(\frac{2x}{1!}+\frac{(2x{)}^3}{3!}+\dots \right)$
- The coefficient of $x^2$ in the above series is $0.$