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 $3^8$?

Hint: Take a look at the proof of why $$\sum_{k=0}^8{8\choose k}$$ is equal to $2^8$. I presume that it was done using the binomial theorem. Using the same theorem, try to modify it so you will get the $2^k$ values in the expansion.
