How to prove that $\sum_{r=0}^n\binom{n}{r}2^r=3^n$
Solution 1:
Hint: Expand $(1+2)^n$ in the usual way.
The result is a special case of a familiar result.
Solution 2:
${n \choose r}$ is the number of $r$-element subsets of $n$-element set.
$\sum_{r=0}^{n}{n \choose r}$ - number of all subsets of $n$-element set, so $\sum_{r=0}^{n}{n \choose r}=2^n$ (why?).
$2^r$ - number of all subsets of $r$-element set.
What we can say now about: $\sum_{r=0}^{n}{n \choose r}2^r$?