Taylor's Polynomial Evaluation Hi guys now started off doing some taylor expansion wanted some help with the following question:
$$ f:[0,\infty)\to \mathbb{R}$$
$$ f(x) = 2e^{-x/2} + e^{-x} $$
Determine the nth degree taylor's polynomila $T_n$ of f about c = 0
My attempt to this question can be seen below;
$$P(x) = f(x) + f'(x)(x-c) + \frac{f''(x)(x-c)^2}{2!} + \frac{f'''(x)(x-c)^3}{3!} + \frac{f''''(x)(x-c)^4}{4!}$$
Evaluation of the function up to the fourth derivative 
$$f'(x) = -e^{-x/2} - e^{-x}$$
$$f''(x) = \frac{1}{2}e^{-x/2} + e^{-x}$$
$$f'''(x) = -\frac{1}{4}e^{-x/2} - e^{-x}$$
$$f''''(x) = \frac{1}{8}e^{-x/2} + e^{-x}$$
substituting x = 0 and we obtain the following
$$f'(x) = -2 $$
$$f''(x) = \frac{3}{2} $$
$$f'''(x) = \frac{-5}{4}$$
$$f''''(x) = \frac{9}{8} $$
substituting c = 0 and the derivatives into the taylor's polynomial we obtain the following:-
$$P(x) = 3 - 2x + \frac{\frac{3}{2}x^2}{2!} - \frac{\frac{5}{4}x^3}{3!} + \frac{\frac{9}{8}x^3}{4!} $$
I am not sure if my evaluation is correct and if so I am having difficulty putting into the form as required below hoping someone can help.
$$T_n (x) = \sum_{k=o}^\infty \frac{f^{k}(c)(x-c)^k}{k!}$$

If you start with the usual
$$e^{-t}=\sum_{n=0}^\infty \frac{(-1)^n}{n!} t^n$$ and apply it twice
$$f(x)=2 e^{-x/2}+e^{-x}=2\sum_{n=0}^\infty \frac{(-1)^n}{2^n\,n!}x^n+\sum_{n=0}^\infty \frac{(-1)^n}{n!} x^n=\sum_{n=0}^\infty \frac{(-1)^n}{n!}\left(1+2^{-n+1} \right)x^n$$
