Finding the generating function of a series with a binomial coefficient and a exponential coefficient

Question

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

Answer 1

$$\sum_{n=0}^\infty \frac{2^8\binom 8n x^n}{2^n}=\sum_{n=0}^8\binom 8n 2^{8-n}x^n =(2+x)^8\qquad \blacksquare$$

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OR

$$(a+x)^N=a^N\left(1+\frac xa\right)^N\\ =a^N\sum_{n=0}^N \binom Nn \left(\frac xa\right)^n\\ =\sum_{n=0}^\infty a^N\binom Nn \left(\frac xa\right)^n$$

Put $a=2, N=8$: $$(2+x)^8=\sum_{n=0}^\infty 2^8\binom 8n \left(\frac x2\right)^n\qquad \blacksquare$$

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Answer 2

Hint:

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