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