Answer
\begin{aligned}
T_{n}(x) =\frac{1}{4}-\frac{1}{4}(x-2)+\frac{3}{16}(x-2)^{2}-\frac{1}{8}(x-2)^{3}+(-1)^{n}(x-2)^{n} \frac{n+1}{2^{n+2}}
\end{aligned}
Work Step by Step
Since
\begin{array}{ll}
{f(x)=x^{-2},} & {f(2)=\frac{1}{4}} \\
{f^{\prime}(x)=-2 x^{-3},} & {f^{\prime}(2)=-2 \cdot \frac{1}{8}} \\
{f^{\prime \prime}(x)=6 x^{-4},} & {f^{\prime \prime}(2)=6 \cdot \frac{1}{16}} \\
{f^{\prime \prime \prime}(x)=-24 x^{-5},} & {f^{\prime \prime \prime}(2)=-24 \cdot \frac{1}{32}} \\
{\vdots} & {\vdots} \\
{f^{(n)}(x)=(-1)^{n} \cdot(n+1) ! \cdot x^{-n-2}} & {f^{(n)}(2)=(-1)^{n} \cdot \frac{(n+1) !}{2^{n+2}}}
\end{array}
Then \begin{aligned}
T_{n}(x) &=f(2)+\frac{f^{\prime}(2)}{1 !}(x-2)+\frac{f^{\prime \prime}(2)}{2 !}(x-2)^{2}+\cdots+(x-2)^{n} \frac{f^{(n)}(2)}{n !} \\
&=\frac{1}{4}-\frac{1}{4}(x-2)+\frac{3}{16}(x-2)^{2}-\frac{1}{8}(x-2)^{3}+(-1)^{n}(x-2)^{n} \frac{n+1}{2^{n+2}}
\end{aligned}
