Answer
-$1+(x-2)-(x-2)^2+(x-2)^3-....$
Work Step by Step
We are given that $f(x)=\dfrac{1}{1-x}$
The Taylor polynomial of order $n$ for the function $f(x)$ at the point $k$ can be defined as:
$p_n(x)=f(k)+\dfrac{f'(k)}{1!}(x-k)+\dfrac{f''(k)}{2!}(x-k)^2+....+\dfrac{f^{n}(k)}{n!}(x-k)^n$
Now, $f'(x)=(1-x)^{-2}\\f''(x)=2(1-x)^{-3}\\f'''(x)=6(1-x)^{-4}$
Here, $f(2)=-1 \\ f'(2)=1 \\f''(2)=-2\\ f'''(2)=6$
Thus, the given series follows the pattern of
$f(x)=\dfrac{1}{1-x}=-1+(x-2)+\dfrac{(-2)}{2!}(x-2)^2+\dfrac{6}{3!}(x-2)^3....=-1+(x-2)-(x-2)^2+(x-2)^3-....$
