Question

The expansion of $\frac{e^{7x}-e^x}{e^{4x}}$ is equal to

• A. $2[1+\frac{(3x{)}^2}{2!}+\frac{(3x{)}^4}{4!}+...]$
• B. $2[\frac{3x}{1!}+\frac{(3x{)}^3}{3!}+...]$
• C. $2[1+\frac{(2x{)}^2}{2!}+\frac{(2x{)}^4}{4!}+...]$
• D. $2[\frac{2x}{1!}+\frac{(2x{)}^3}{3!}+...]$

Answer 1

The correct option is B $2[\frac{3x}{1!}+\frac{(3x{)}^3}{3!}+...]$

$\frac{e^{7x}-e^x}{e^{4x}}$
$=e^{3x}-e^{-3x}=[1+\frac{3x}{1!}+\frac{(3x{)}^2}{2!}+...]-[1-\frac{3x}{1!}+\frac{(3x{)}^2}{2!}-...]=2[\frac{3x}{1!}+\frac{(3x{)}^3}{3!}+...]$