TITLE: Evaluate $\sum\limits_{k=1}^n k^2$ and $\sum\limits_{k=1}^n k(k+1)$ combinatorially
QUESTION [22 upvotes]: $$\text{Evaluate } \sum_{k=1}^n k^2 \text{ and } \sum_{k=1}^{n}k(k+1) \text{ combinatorially.}$$

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 $r=2$

REPLY [6 votes]: The story of André can be continued:
Bob was unhappy that he always had the least chairs to choose from. So Alice and Bob decided to choose their chair at the same time. If both choose the same chair they shared the chair. So if Lefty chose the second seat from right Alice and Bob had to select the first seat. Lefty could select the third seat leaving $2$ possibilities to Alice and $2$ possibilities to Bob. Lefty could select the leftmost chair (chair with number $n+1$) giving them both $n$ possibilities to choose. So the total number of arrangements was now 
$$\sum_{k=1}^n k^2$$
Again we can calculate the possibilities in another way: Select 3 chairs from $n+1$ chairs, the leftmost is for Lefty the remaining two can be assigned to Alice and Bob. This is the case when Alice and Bob choose different seats and the number is 
$$2\binom{n+1}{3}$$
When we select only 2 from the $n+1$ chairs, the leftmost is for Lefty and the second must be shared by Alice and Bob. This gives 
$$\binom{n+1}{2}$$
possibilities. The number of all possible selections is the sum of both numbers that is
$$2\binom{n+1}{3}+\binom{n+1}{2} = n(n+1)(2n+1)/6$$