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)

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