Answer
$\dfrac{1}{4}-\dfrac{1}{4^2}(x-3)+\dfrac{1}{4^3}(x-3)^2-\dfrac{1}{4^4}(x-3)^3....$
Work Step by Step
We are given that $f(x)=\dfrac{1}{x+1}$
The 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, $f'(x)=(x+1)^{-1}\\f''(x)=2(x+1)^{-3}\\f'''(x)=-6(x+1)^{-4}$
Here, $f(3)=\dfrac{1}{4} \\ f'(3)=-\dfrac{1}{(4)^2} \\f''(3)=\dfrac{2}{(4)^3}\\ f'''(3)=-\dfrac{6}{(4)^4}$
Thus, the given series follows the pattern of
$f(x)=\dfrac{1}{x+1}=\dfrac{1}{4}+(-\dfrac{1}{(4)^2})(x-3)+\dfrac{\dfrac{2}{(4)^3}}{2!}(x-3)^2+\dfrac{(-\dfrac{6}{(4)^4})}{3!}(x-3)^3....=\dfrac{1}{4}-\dfrac{1}{4^2}(x-3)+\dfrac{1}{4^3}(x-3)^2-\dfrac{1}{4^4}(x-3)^3....$
