Answer
$\Sigma_{n=0}^{\infty} (-1)^{n+1} (x-2)^n$
Work Step by Step
Here, we have: $f(x)=\dfrac{1}{1-x}\implies f(2)=-1\\ f'(x)=\dfrac{1}{(x-2)^2} \implies f'(2)=1\\f''(x)=\dfrac{2}{(1-x)^3} \implies f''(2)=-2$
and so on.
The Taylor series of a function $f(x)$ at $x=c$ can be written as:
$T(x)=f(a)+\dfrac{f′(a)}{1!}(x−a)+\dfrac{f′′(a)}{2!}(x−a)^2+......+\dfrac{f^n(a)}{n!}(x−a)^n$
Hence, the required Taylor series is:
$T(x)=-1+1(x-2)+\dfrac{-2}{2!}(x-2)^2+.......=\Sigma_{n=0}^{\infty} (-1)^{n+1} (x-2)^n$
