combinatorics - Combinatorial interpretation of sum of squares, cubes

Consider the sum of the first $n$ integers:

$$\sum_{i=1}^n\,i=\frac{n(n+1)}{2}=\binom{n+1}{2}$$
This has always made the following bit of combinatorial sense to me. Imagine the set $\{*,1,2,\ldots,n\}$. We can choose two from this set, order them in decreasing order and thereby obtain a point in $\mathbb{N}^2$. We interpret $(i,*)$ as $(i,i)$. These points give a clear graphical representation of $1+2+\cdots+n$:

$$\begin{matrix} &&&\circ\\ &&\circ&\circ\\ &\circ&\circ&\circ\\ \circ&\circ&\circ&\circ\\ \end{matrix}$$

Similar identities are:

$$\sum_{i=1}^n\,i^2=\frac{n(n+1)(2n+1)}{6}=\frac{2n(2n+2)(2n+1)}{24}=\frac{1}{4}\binom{2n+2}{3}$$
$$\sum_{i=1}^n\,i^3=\frac{n^2(n+1)^2}{4}=\binom{n+1}{2}^2$$

I am aware of geometric explanations of these identities, but not combinatorial ones similar to the above explanation for summing first powers that make direct use of the "choosing" interpretation of the binomial coefficient. Can anyone offer combinatorial proofs of these?

Answer

Here's a combinatorial proof for

$$\sum_{k=1}^n k^2 = \binom{n+1}{2} + 2 \binom{n+1}{3},$$ which is just another way of expressing the sum. Both sides count the number of ordered triples $(i,j,k)$ with $0 \leq i,j < k \leq n$.

For the left side, condition on the value of $k$. For each $k$, there are $k^2$ ways to choose $i$ and $j$ from the the set $\{0, 1, \ldots, k-1\}$.

For the right side, consider the cases $i=j$ and $i \neq j$ separately. If $i = j$, then there are $\binom{n+1}{2}$ such triples. This is because we just choose two numbers from $\{0, \ldots, n\}$; the smaller must be the value of $i$ and $j$ and the larger must be the value of $k$. If $i \neq j$, then there are $2\binom{n+1}{3}$ such triples, as we could have $i < j$ or $j < i$ for the smaller two numbers.

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For

$$\sum_{k=1}^n k^3 = \binom{n+1}{2}^2,$$
both sides count the number of ordered 4-tuples $(h,i,j,k)$ with $0 \leq h,i,j < k \leq n$.

For the left side, once again if we condition on the value of $k$ we see that there are $\sum_{k=1}^n k^3$ such 4-tuples.

For the right side, there is a bijection from these 4-tuples to ordered pairs of two-tuples $(x_1,x_2), (x_3,x_4)$ with $0 \leq x_1 < x_2 \leq n$ and $0 \leq x_3 < x_4 \leq n$. There are $\binom{n+1}{2}^2$ such pairs, so let's look at the bijection.

The bijection: If $h < i$, then map $(h,i,j,k)$ to $(h,i),(j,k)$. If $h > i$, then map $(h,i,j,k)$ to $(j,k), (i,h)$. If $h = i$, then map $(h,i,j,k)$ to $(i,k), (j,k)$. This mapping is reversible, as these three cases correspond to the cases where $x_2 < x_4$, $x_2 > x_4$, and $x_2 = x_4$.

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(Both of these proofs are in Chapter 8 of Proofs that Really Count, by Benjamin and Quinn. They give at least one other combinatorial proof for each of these identities as well.)