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