How to interpret the results $$ 1^2+2^2+\ldots+n^2=\binom{n+1}{2}+2\binom{n+1}{3} \\ 1^3+2^3+\ldots+n^3=\binom{n+1}{2}+6\binom{n+1}{3}+6\binom{n+1}{4} $$
I want to find a clear argument (combinatorial example,etc.) to prove this, other than induction or merely use the formula.
The sum of cubes formula can also be derived by slightly generalizing this answer by Mike Earnest for the sum of squares case.
Consider counting ordered quadruplets of integers $(w,x,y,z)$ s.t.
$0 \le w, x, y < z$
$1 \le z \le n$
When $z=k$ the number of ways to choose $(w,x,y)\in \{0, 1, \dots, k-1\}^3$ is $k^3$, so the total is $\sum_{k=1}^n k^3$.
OTOH all such quadruplets can be classified as follows:
$w, x, y$ all distinct: For a particular ordering, say $w < x < y < z$, we have ${n+1 \choose 4}$ choices. There are $3!=6$ orderings among the $\{w,x,y\}$, so this case gives $6 \times {n+1 \choose 4}$.
$w = x = y$: Then we're simply choosing two numbers, $w$ and $z$. So this case gives ${n+1 \choose 2}$.
Exactly two of the three $\{w, x, y\}$ are equal: There are $3$ ways to choose the equal pair, then $2$ ways to decide if the remaining value is bigger or smaller. Once we decided that, say $x < w = y$, we are simply picking three numbers $x, w, z$. So this case gives $3 \times 2 \times {n+1 \choose 3}$.
Two different combinatorial interpretations of your first identity are given in the answers to this question. My answer (not the accepted one) was based on counting the number of ways you can place two white bishops on an $n\times n$ chessboard so that they guard each other. Calling this number $B_n$, and counting in two different ways, I showed that $$B_n=2\binom{n+1}3+2\binom n3$$ and $$B_n=2[1^2+2^2+\cdots+(n-1)^2].$$ For details see my answer to that old question.
