Answer
$f(x)=1+2x+2x^2+\dfrac{4}{3}x^3+\dfrac{2}{3}x^4+......$
Work Step by Step
We are given the function:
$f(x)=e^{2x}=(e^2)^x$.
Determine the Maclaurin series corresponding to $f(x)$:
$f(x)=f(0)+\dfrac{f'(0)}{1!}+\dfrac{f"(0)}{2!}+.....$
$f'(x)=2e^{2x}\Rightarrow f'(0)=2$
$f"(x)=2\cdot2e^{2x}=2^2e^{2x}\Rightarrow f"(0)=2^2$
$f^{(3)}(x)=2\cdot2^2e^{2x}=2^3e^{2x}\Rightarrow f^{(3)}(0)=2^3$
....
$f^{(n)}(x)=2^n e^{2x}\Rightarrow f^{(3)}(0)=2^n$
$f(x)=1+\dfrac{2}{1!}x+\dfrac{2^2}{2!}x^2+\dfrac{2^3}{3!}x^3+\dfrac{2^4}{4!}x^3+.....$
$f(x)=1+2x+2x^2+\dfrac{4}{3}x^3+\dfrac{2}{3}x^4+......$
