TITLE: Prove $\sum\binom{n}{k}2^k = 3^n$ using the binomial theorem
QUESTION [12 upvotes]: I'm studying for a midterm and need some help with proving summation

$$\sum\limits_{k=0}^n\binom{n}{k}2^k = 3^n$$

using the binomial theorem.
This is what I've been thinking so far: 
In the binomial theorem, we set $x = 0$ and $y = 2$, so:
$3^n = (x+y)^n = \sum\limits_{k=0}^n\binom{n}{k}y^k$
$ = \sum\limits_{k=0}^n\binom{n}{k}2^k$
Am I getting this correct so far or completely wrong?
Any help would be appreciated. Thanks.

REPLY [3 votes]: You are almost there - the value of $x$ should be 1 rather than 2, so that we get (essentially what you already have written):
$$3^n = (1+2)^n = \sum_{k=0}^n \binom{n}{k} 1^{n-k} 2^k = \sum_{k=0}^n \binom{n}{k} 2^k.$$