Prove the following binomial identity using a counting proof:
$$\sum\limits_{r=0}^n (r+1)\cdot {n \choose r} = (n+2)\cdot 2^{n-1}$$
I expanded the right side to get: $2^n + n2^{n-1}$, because I think that will be easier to count but have not been able to actually get any farther. Thank you.
On
You started correctly. Then your equality is equivalent to $\sum_0^n {n\choose r}+\sum_0^n r {n\choose r}=2^n+n2^{n-1}$. But by the binomial formula $\sum_0^n {n\choose r}=2^n$. So your equality is equivalent to $\sum_0^n r {n\choose r}=n2^{n-1}$. This can be proved by induction on $n$ or by counting chaired committees as in @BarryCipra's comment. This has been done here already.
On
Expand the lefthand side as well:
$$\sum_{r=0}^n(r+1)\binom{n}r=\sum_{r=0}^nr\binom{n}r+\sum_{r=0}^n\binom{n}r\,.$$
The second summation gives the number of ways to choose a team of any size from a pool of $n$ players; clearly this is just the number of subsets of the pool, which is $2^n$. The first summation gives the number of ways to choose a team of any size $r$ and then appoint one of those $r$ players captain; that can just as well be done by choosing the captain and then choosing $r-1$ players from the remaining $n-1$ to fill out the team, something that can be done in $n\binom{n-1}{r-1}$ ways. Thus,
$$\sum_{r=0}^nr\binom{n}r=n\sum_{r=0}^n\binom{n-1}{r-1}=n\sum_{r=0}^{n-1}\binom{n-1}{r-1}\,,$$
which is just $n$ times the number of ways to pick a subset of the remaining $n-1$ players, or $n2^{n-1}$. And this is exactly the righthand side that you have.
(By the way, the identity $r\binom{n}r=n\binom{n-1}{r-1}$ is often useful and is worth knowing.)
If only counting is allowed, consider the number of all committees, chaired and not.
We can count them in two ways. First choose number $r=0,...,n$ and a subset with $r$ elements from a set of $n$ elements. Then we choose a chair of that committee or no chair at all. The total number of choices is the LHS of your equality.
We can compute it in a different way. First count the number of all chairless committees. These are all $2^n$ subsets of the $n$-element set. Then we count the chaired committees by first choosing a chair ($n$ choices) and then choosing a subset of the set of members witbout the chosen chair ($2^{n-1}$ choices). Altogether the number is equal to the RHS of your equality.
So LHS=RHS.