Reason what 2^k in a sigma will do I am trying to solve the following calculation without a calculator:
$$\sum_{k=0}^82^k{8\choose k}$$
The first part: 
$$\sum_{k=0}^8{8\choose k}$$
is equal to $2^8$. I already know that the answer will be $3^8$. How did the $2^k$ transform the answer from $2^8$ to $3^8$?

Use that $$
 (a+b)^n = \sum_{k=0}^n a^k b^{n-k} \binom{n}{k} \text{.}
$$
For $a=2$, $b=1$ you get the sum you want to compute, and the result is therefore $(2+1)^8$
