Finding the generating function of a series with a binomial coefficient and a exponential coefficient So I am given this series
$$2^8, 2^7 \binom{8}{1}, 2^6 \binom{8}{2}, 2^5 \binom{8}{3}, 2^4 \binom{8}{4}, 2^3 \binom{8}{5}, 2^2 \binom{8}{6}, 2^1 \binom{8}{7}, \binom{8}{8}, 0, 0, 0, 0, ...$$
which I converted to the summation 
$$\sum_{n=0}^\infty \frac{(2^8 \cdot \binom{8}{n} \cdot x^n)}{2^n}  $$
The problem is to find the closed form generating function for this series
I know that the closed form generating function for 
$$\sum_{n=0}^\infty \binom{8}{n}x^n = (1+x)^8$$ and 
$$\sum_{n=0}^\infty \frac{(2^8 \cdot x^n)}{2^n} = \frac{2^8}{1-x/2}$$
I just can't find anything in my book or online on how to deal with the combination of the two 

Hint:
$$ \sum \frac{2^8 \binom{8}{n} x^n}{2^n} = \sum 2^8 \binom{8}{n} \left(\frac x2\right)^n = \cdots  $$
