Prove $\sum^n_{k=0}{n \choose k} 2^k=3^n$

Question

I need to prove that $\sum^n_{k=0}{n \choose k} 2^k=3^n$

I already know that $\sum^n_{k=0}{n \choose k}=2^n$

I'm not really sure where to go after this.

Answer 1

Hint: $$3^n=(1+2)^n$$

And binomial theorem.

Answer 2

For a combinatorial argument, $3^n$ is the number of ternary strings of length $n$. Each such string has some number $0\leqslant k\leqslant n$ of digits equal to $0$ or $1$. There are $2^k$ binary strings of length $k$, and $\binom nk$ ways each binary string may appear within a ternary string of length $n$. Therefore $$\sum_{k=0}^n \binom nk 2^k $$ also counts the number of ternary strings of length $n$.