Answer
$e^2+e^2(x-2)+e^2\dfrac{(x-2)^2}{2!}+.... $
or,
$\Sigma_{n=0}^{\infty} e^2\dfrac{(x-2)^n}{n!}$
Work Step by Step
Differentiate the given function $f(x)$ as follows:
$f'(x)=e^x \implies f'(2)=e^2 \\f''(x)=e^x \implies f''(0)=3\\ f'''(x)=e^2 \implies f'''(0)=e^2$
The Taylor series at $x=2$ can be written as:
$f(x)=f(2)+f'(2) (x-2)+\dfrac{f''(2)(x-2)^2 }{2!}+....\\=e^2+e^2(x-2)+e^2\dfrac{(x-2)^2}{2!}+.... \\=\Sigma_{n=0}^{\infty} e^2\dfrac{(x-2)^n}{n!}$
