Answer
$\Sigma_{k=0}^{\infty} \dfrac{(-1)^k}{5^{k+1}}(x-3)^k$
Work Step by Step
The Maclaurin Series is
$ \displaystyle\sum\limits_{n=0}^{\infty} \frac{f^n(0)x^n} {n!} = f(0) + f'(0)x + \frac{(f''(0)x^2)}{2!} + ..\\
$
$f(x)=\dfrac{1}{x+2}$
Differentiate
$f'(x)=\dfrac{-1}{(x+2)^2}$
$f''(x)=\dfrac{2}{(x+2)^3}$
$f'''(x)=\dfrac{-6}{(x+2)^4}$
Now $f(3)=\dfrac{1}{5}$ , $f'(3)=-\dfrac{1}{25}$ , $f''(3)=\dfrac{2}{125}$, $f'''(3)=\dfrac{-6}{625}$
Hence the Maclaurin Series for the function is
$\displaystyle \dfrac{1}{5}-\dfrac{1}{25}(x-3) +\dfrac{1}{125}(x-3)^3 .......+ \dfrac{(-1)^n}{5^{n+1}}(x-3)^k =\Sigma_{k=0}^{\infty} \dfrac{(-1)^k}{5^{k+1}}(x-3)^k$
