Answer
$\Sigma_{k=0}^{n} (-1)^k x^k$
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!} + ..\\
$
$f(x)=\dfrac{1}{1+x}$
Differentiate w.r.t x
$f'(x)=-\dfrac{1}{(1+x)^2}$
$f''(x)=\dfrac{2}{(1+x)^3}$
$f'''(x)=\dfrac{-6}{(1+x)^4}$
$f''''(x)=\dfrac{24}{(1+x)^5}$
Now $f(0)=1$, $f'(0)=-1$, $f''(0)=2$, $f'''(0)=-6$ and $f''''(0)=24$
Hence the Maclaurin Series for the function is
$1-x+x^2-x^3+ .......+(-1)^n x^n=\Sigma_{k=0}^{n} (-1)^k x^k$
