Evaluate $\sum\limits_{k=1}^n k^2$ and $\sum\limits_{k=1}^n k(k+1)$ combinatorially 
$$\text{Evaluate } \sum_{k=1}^n k^2 \text{ and } \sum_{k=1}^{n}k(k+1) \text{ combinatorially.}$$

For the first one, I was able to express $k^2$ in terms of the binomial coefficients by considering a set $X$ of cardinality $2k$ and partitioning it into two subsets $A$ and $B$, each with cardinality $k$. Then, the number of ways of choosing 2-element subsets of $X$ is $$\binom{2k}{2} = 2\binom{k}{2}+k^2$$ So sum $$\sum_{k=1}^n k^2 =\sum_{k=1}^n \binom{2k}{2} -2\sum_{k=2}^n \binom{k}{2} $$ $$ \qquad\qquad = \color{red}{\sum_{k=1}^n \binom{2k}{2}} - 2 \binom{n+1}{3} $$ I am stuck at this point to evaluate the first of the sums. How to evaluate it?
I need to find a similar expression for $k(k+1)$ for the second sum highlighted above. I have been unsuccessful this far.  (If the previous problem is done then so is this, but it would be nice to know if there are better approaches or identities that can be used.)
Update: I got the second one. Consider $$\displaystyle \binom{n+1}{r+1} = \binom{n}{r}+\binom{n-1}{r}+\cdots + \binom{r}{r}$$ Can be shown using recursive definition. Now multiply by $r!$ and set $r=2$

For 
$$\sum_{k=1}^n (k+1)(k),$$
let me tell a story.
There are $n+2$ chairs in a row, and three people, Lefty, Alice, and Bob.
They want to sit down.  Lefty does not like to have anyone to his left.  In how many ways can they sit?
Lefty could be in the third seat from the right, leaving $2$ choices for Alice and then $1$ for Bob.  Or else Lefty could be in the fourth seat from the right, leaving $3$ choices for Alice and then $2$ for Bob. And so on.  Finally Lefty could be in the leftmost seat, giving Alice $n+1$ choices and Bob $n$.  So the total number of seating arrangements is our sum.
Or else choose $3$ seats from $n+2$, reserve the leftmost for Lefty.  For each choice of $3$ seats there are then $2$ places for Alice to sit, for a total of 
$$2\binom{n+2}{3}.$$
