<s>[INST] interpretation on $1^2+2^2+\ldots+n^2=\binom{n+1}{2}+2\binom{n+1}{3}$ and $1^3+2^3+\ldots+n^3=\binom{n+1}{2}+6\binom{n+1}{3}+6\binom{n+1}{4}$ 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.
 [/INST] 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 $3 \times 2 \times {n+1 \choose 3}$.
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