Answer
$A_{0}(x)=1 \\ A_{1}(x)=1+2x \\A_{2}(x)=1+2x+2x^2 \\A_{3}(x)=1+2x+2x^2+\dfrac{4}{3}x^3$
Work Step by Step
The Taylor polynomial is defined for the function $f(x)$ at the point $a$ of order $n$ as:
$A_n(x)=f(a)+\dfrac{f'(a)}{1!}(x-a)+\dfrac{f''(a)}{2!}(x-a)^2+....+\dfrac{f^{n}(a)}{n!}(x-a)^n$
Now, we get $f(0)=1 \\ f'(x)=e^{2x} \implies f'(0)=2 \\ f''(x)=2e^{2x} \implies f''(0)=4 \\ f'''(x)=8e^{2x} \implies f'''(0)=8$
Taylor polynomial is defined as: $A_{0}(x)=1 \\ A_{1}(x)=1+2x \\A_{2}(x)=1+2x+2x^2 \\A_{3}(x)=\dfrac{1}{2}+(\dfrac{-1}{4})(x-0)+(\dfrac{1}{4})\dfrac{(x-0)^2}{2!}+(\dfrac{-3}{8})\dfrac{(x-0)^3}{3!}=1+2x+(\dfrac{4}{2})x^2+(\dfrac{8}{6})x^3=1+2x+2x^2+\dfrac{4}{3}x^3$
