Question

The 4th term in the expansion of $\frac{e^{4x}-1}{2e^{2x}}$ is

• A. $\frac{(2x{)}^7}{7!}$
• B. $\frac{(2x{)}^4}{4!}$
• C. $\frac{(2x{)}^5}{5!}$
• D. $\frac{(2x{)}^6}{6!}$

Answer 1

The correct option is A $\frac{(2x{)}^7}{7!}$

$=\frac{e^{2x}-e^{-2x}}{2}$

using $e^{2x}={\sum }_{n=0}^{\infty }\frac{(2x{)}^k}{k!}$ and $e^{-2x}={\sum }_{n=0}^{\infty }\frac{(-2x{)}^k}{k!}$ in the given equation :

We get :
$=\frac{e^{2x}-e^{-2x}}{2}$ $={\sum }_{n=1}^{\infty }\frac{(2x{)}^{2n-1}}{(2n-1)!}$
So $4^{th}$ term $=\frac{(2x{)}^7}{7!}$