Answer
$\Sigma_{n=0}^{\infty} (-1)^{n} \dfrac{(4x)^n}{n !}$
Work Step by Step
Here, we have: $f(x)=e^{-4x} \implies f(0)=1\\ f'(x)=-4 e^{-4x} \implies f'(0)=-4\\f''(x)=16 e^{-4x} \implies f''(0)=16$
and so on.
The Taylor series of a function $f(x)$ at $x=c$ can be written as:
$T(x)=f(a)+\dfrac{f′(a)}{1!}(x−a)+\dfrac{f′′(a)}{2!}(x−a)^2+......+\dfrac{f^n(a)}{n!}(x−a)^n$
Hence, the required Taylor series is:
$T(x)=1-4x+\dfrac{16x^2}{2!}+.......=\Sigma_{n=0}^{\infty} (-1)^{n} \dfrac{(4x)^n}{n !}$
