Question

The coefficient of $ x^3 $ in the expansion of $ e^{2x+3} $ as a series in powers of $ x $ is

• A. $ e^3 $
• B. $ \frac{3}{4}e^3 $
• C. $ \frac{4}{3}e^3 $
• D. $None\, of\, these$

Answer 1

Answer: (A)

Given, $f \left(x\right)=e^{2x+3}$
$f'\left(x\right)=2\cdot e^{2x+3}$, $f'\left(0\right)=2\cdot e^{3}$
$f''\left(x\right)=4\cdot e^{2x+3}$, $f''\left(0\right)=4\cdot e^{3}$
$f''\left(x\right)=8\cdot e^{2x+3}$, $f''\left(0\right)=8\cdot e^{3}$
By Maclaurin’s series,
$f \left(e^{2x+3}\right)=1+\frac{x\cdot f'\left(0\right)}{1!}+\frac{x^{2}f''\left(0\right)}{2!}+\frac{x^{3}f''\left(0\right)}{3!}+\ldots$
$=1+\frac{x\cdot\left(2e^{3}\right)}{1!}+\frac{x^{2}\left(4e^{3}\right)}{2!}+\frac{x^{3}\cdot\left(8e^{3}\right)}{3!}+\ldots$
Hence, the coefficient $x^{3}$ in the expansion of $e^{2x+3}$
$=\frac{8e^{3}}{3!}=\frac{8e^{3}}{6}=\frac{4}{3}e^{3}$