Answer
$\Sigma_{k=0}^{n} \frac {(-1)^kx^k}{k!}$
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!} + ..\\
$
$f(x)=e^{-x}$
Differentiate w.r.t $x$.
$f'(x)=-e^{-x}$
$f''(x)=e^{-x}$
$f'''(x)=-e^{-x}$
$f''''(x)=e^{-x}$
Now $f(0)=1$ , $f'(0)=-1$ , $f''(0)=1$, $f'''(0)=-1 $ and $f''''(0)=1$
Hence, the Maclaurin Series for the function is
$\displaystyle \pi 1 -x+ \frac {x^2} {2} + .......+ \frac {(-1)^nx^{n} } {n!} = \Sigma_{k=0}^{n} \frac {(-1)^kx^k}{k!}$
