Answer
$-1+(x-2)-(x-2)^2+(x-2)^3-....$
Work Step by Step
Given: $f(x)=\dfrac{1}{1-x}$
Taylor polynomial of order $n$ for the function $f(x)$ at the point $a$ can be defined as:
$A_n(x)=f(k)+\dfrac{f'(k)}{1!}(x-a)+\dfrac{f''(a)}{2!}(x-a)^2+....+\dfrac{f^{n}(a)}{n!}(x-a)^n$
Now, $f(2)=-1$ and $f'(x)=(1-x)^{-2} \implies f'(2)=1\\f''(x)=2(1-x)^{-3} \implies f''(2)=-2\\f'''(x)=6(1-x)^{-4} \implies f'''(2)=6$
Thus, we have $f(x)=-1+(x-2)+\dfrac{(-2)(x-2)^2}{2!}+\dfrac{6(x-2)^3}{3!}....$
and $f(x)=-1+(x-2)-(x-2)^2+(x-2)^3-....$
