Answer is: option4
\( \frac{4}{3} \)Solution:
From the well known expansion of the exponential function: \[ e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots \]
We have \[ e^{2x}=1+(2x)+\frac{(2x)^2}{2!}+\frac{(2x)^3}{3!}+\cdots \]
Then the coefficient of \(x^3\) is \[ \frac{2^3}{3!}=\frac{8}{6}=\frac{4}{3} \]
Solution II.
If \(f(x)=e^{2x}\), then the coefficient of \(x^3\) will be \[ \frac{f^{(3)}(0)}{3!} \]
\[ f(x)=e^{2x} \] \[ f'(x)=2e^{2x},\qquad f''(x)=4e^{2x},\qquad f^{(3)}(x)=8e^{2x} \]
\[ f^{(3)}(0)=8 \]
Therefore the coefficient of \(x^3\) is \[ \frac{8}{3!}=\frac{8}{6}=\frac{4}{3} \]
