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