Answer
$2+\Sigma_{n=1}^\infty 3x^n$
Work Step by Step
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, $f'(x)=3(1-x)^{-2}\\f''(x)=6(1-x)^{-3}\\f'''(x)=18(1-x)^{-4}$
and $f^{n}(x)=3(n!)(1-x)^{-n-1}$
Here, $f(0)=2 \\ f'(0)=3=3 \cdot 1! \\f''(0)=6=3 \cdot 2!\\ f'''(0)=18=3 \cdot 3 !$
Thus, the given series follows the pattern of
$2+3x+3x^2+3x^3+....=2+\Sigma_{n=1}^\infty 3x^n$
