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. $Noneofthese$

Answer 1

Answer: (A)

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