Answer: \((r+1)^4-r^4=4r^3+6r^2+4r+1\).
Summing from \(r=1\) to \(r=n\) gives
\((n+1)^4-1=4\sum_{r=1}^{n}r^3+6\sum_{r=1}^{n}r^2+4\sum_{r=1}^{n}r+n.\)
Using \(\sum r=\frac{n(n+1)}{2}\) and \(\sum r^2=\frac{n(n+1)(2n+1)}{6}\), this simplifies to
\(4\sum_{r=1}^{n}r^3=n^2(n+1)^2.\)
Therefore
\(\sum_{r=1}^{n}r^3=\frac14n^2(n+1)^2.\)
Expand the fourth power first:
\((r+1)^4=r^4+4r^3+6r^2+4r+1\), so
\((r+1)^4-r^4=4r^3+6r^2+4r+1\).
Now sum both sides from \(r=1\) to \(r=n\):
\(\sum_{r=1}^{n}\big((r+1)^4-r^4\big)=\sum_{r=1}^{n}(4r^3+6r^2+4r+1)\).
The left side telescopes:
\((2^4-1^4)+(3^4-2^4)+\cdots+((n+1)^4-n^4)=(n+1)^4-1^4\).
Therefore
\((n+1)^4-1=4\sum_{r=1}^{n}r^3+6\sum_{r=1}^{n}r^2+4\sum_{r=1}^{n}r+n\).
Use the standard formulas
\(\sum_{r=1}^{n}r=\frac{n(n+1)}{2}\), \(\sum_{r=1}^{n}r^2=\frac{n(n+1)(2n+1)}{6}\).
Substitute these in:
\((n+1)^4-1=4\sum_{r=1}^{n}r^3+n(n+1)(2n+1)+2n(n+1)+n\).
Now simplify the right-hand extra terms:
\(n(n+1)(2n+1)=2n^3+3n^2+n\),
\(2n(n+1)=2n^2+2n\).
So
\((n+1)^4-1=4\sum_{r=1}^{n}r^3+2n^3+5n^2+4n\).
Also
\((n+1)^4-1=n^4+4n^3+6n^2+4n\).
Hence
\(n^4+4n^3+6n^2+4n=4\sum_{r=1}^{n}r^3+2n^3+5n^2+4n\).
Rearranging gives
\(4\sum_{r=1}^{n}r^3=n^4+2n^3+n^2=n^2(n^2+2n+1)=n^2(n+1)^2\).
Therefore
\(\sum_{r=1}^{n}r^3=\frac14 n^2(n+1)^2\).
