Answer
$\Sigma_{k=0}^n \dfrac{(-1)^k}{5^{k+1}}(x-3)^n$
Work Step by Step
The nth Taylor polynomial for a function $f$ is
$$p_x(x)= f(x_0) + f'(x_0)(x-x_0) +\dfrac{f''(x_0)}{2!} (x-x_0)^2 +.......+\dfrac{f^n(x_0)}{n!} (x-x_0)^n$$
Given function, $f(x)=\dfrac{1}{x+2}$
Differentiate w.r.t x
$f'(x)=-\dfrac{1}{(x+2)^2}$
$f''(x)=\dfrac{2}{(x+2)^3}$
$f'''(x)=-\dfrac{6}{(x+2)^4}$
$f''''(x)=\dfrac{24}{(x+2)^5}$
Now $f(3)=\dfrac{1}{5}$, $f'(3)=-\dfrac{1}{25}$, $f''(3)=\dfrac{1}{125}$, $f'''(3)=\dfrac{-1}{625}$ and $f''''(3)=\dfrac{1}{3125}$
Hence the Maclaurin Series for the function is
$\displaystyle \dfrac{1}{5}-\dfrac{1}{25}(x-3)+\dfrac{1}{125}(x-3)^2+ .......+\dfrac{(-1)^n}{5^{n+1}}(x-3)^n=\Sigma_{k=0}^n \dfrac{(-1)^k}{5^{k+1}}(x-3)^n$
