Answer
$\dfrac{1}{4}-\dfrac{(x-3)}{4^2}+\dfrac{(x-3)^2}{4^3}-\dfrac{(x-3)^3}{4^4}...$
Work Step by Step
Given: $f(x)=\dfrac{1}{x+1}$
Taylor polynomial of order $n$ for the function $f(x)$ at the point $a$ can be defined as:
$A_n(x)=f(k)+\dfrac{f'(k)}{1!}(x-a)+\dfrac{f''(a)}{2!}(x-a)^2+....+\dfrac{f^{n}(a)}{n!}(x-a)^n$
Now, $f(3)=\dfrac{1}{4}$ and $f'(x)=(x+1)^{-1} \implies f'(3)=-\dfrac{1}{(4)^2}\\f''(x)=2(x+1)^{-3} \implies f''(3)=\dfrac{2}{(4)^3}\\f'''(x)=-6(x+1)^{-4} \implies f'''(3)=-\dfrac{6}{(4)^4}$
Taylor polynomial of order $n$ for the function $f(x)$ is:
$f(x)=\dfrac{1}{4}+(-\dfrac{1}{(4)^2})(x-3)+\dfrac{\dfrac{2}{(4)^3}}{2!}(x-3)^2+\dfrac{(-\dfrac{6}{(4)^4})}{3!}(x-3)^3....$
Hence, $f(x)=\dfrac{1}{4}-\dfrac{(x-3)}{4^2}+\dfrac{(x-3)^2}{4^3}-\dfrac{(x-3)^3}{4^4}...$
