Answer
$f(x)=e^2+e^2(x-2)+e^2\dfrac{(x-2)^2}{2!}+....=\Sigma_{n=0}^{\infty} e^2\dfrac{(x-2)^n}{n!}$
Work Step by Step
We have $f'(x)=e^x \implies f'(2)=e^2$;
$f''(x)=e^x \implies f''(0)=3\\ f'''(x)=e^2 \implies f'''(0)=e^2$ and $f^{4}(1)=e^2$ and so on.
Therefore, the Taylor's series at $x=2$ is as follows:
$f(x)=f(2)+f'(2) (x-2)+\dfrac{f''(2)(x-2)^2 }{2!}+.....so \space on$
or, $f(x)=e^2+e^2(x-2)+e^2\dfrac{(x-2)^2}{2!}+....=\Sigma_{n=0}^{\infty} e^2\dfrac{(x-2)^n}{n!}$