#### Answer

\begin{aligned}
T_{n}(x) =1-x+x^{2}-x^{3}+\cdots+(-1)^{n} x^{n}
\end{aligned}

#### Work Step by Step

Since
\begin{array}{ll}
{f(x)=\frac{1}{1+x},} & {f(0)=1} \\
{f^{\prime}(x)=-\frac{1}{(1+x)^{2}},} & {f^{\prime}(0)=-1} \\
{f^{\prime \prime}(x)=\frac{2}{(1+x)^{3}},} & {f^{\prime \prime}(0)=2} \\
{f^{\prime \prime \prime}(x)=-\frac{6}{(1+x)^{4}},} & {f^{\prime \prime \prime}(0)=-6}
\end{array}
Then \begin{aligned}
T_{n}(x) &=f(0)+\frac{f^{\prime}(0)}{1 !}(x-0)+\frac{f^{\prime \prime}(0)}{2 !}(x-0)^{2}+\cdots+x^{n} \frac{f^{(n)}(0)}{n !} \\
&=1-x+x^{2}-x^{3}+\cdots+(-1)^{n} x^{n}
\end{aligned}