<s>[INST] 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 
 [/INST] $$\sum_{n=0}^\infty \frac{2^8\binom 8n x^n}{2^n}=\sum_{n=0}^8\binom 8n 2^{8-n}x^n
=(2+x)^8\qquad \blacksquare$$

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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