Answer
$\Sigma_{k=0}^{\infty} \frac {(-1)^k(x-\ln 2)^k}{2k!} $
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
$f'(x)=-e^{-x}$
$f''(x)=e^{-x}$
$f'''(x)=-e^{-x}$
Now $f(\ln 2)=\dfrac{1}{2}$ , $f'(\ln 2)=-\dfrac{1}{2}$ , $f''(\ln 2)=\dfrac{1}{2}$, $f'''(\ln 2)=-\dfrac{1}{2}$
Hence the Maclaurin Series for the function is
$\displaystyle \dfrac{1}{2}-\dfrac{1}{2}(x-\ln 2) +\dfrac{1}{4}(x-\ln 2)^2- .......+ \frac {(-1)^n(x-\ln 2)^n}{2n!} = \Sigma_{k=0}^{\infty} \frac {(-1)^k(x-\ln 2)^k}{2k!} $
