Answer: (a) \((r+1)^4-r^4=4r^3+6r^2+4r+1\).
(b) \(\displaystyle \sum_{r=1}^{n}r^3=\frac{1}{4}n^2(n+1)^2\).
Expand using the binomial theorem:
\((r+1)^4=r^4+4r^3+6r^2+4r+1\).
Therefore
\((r+1)^4-r^4=4r^3+6r^2+4r+1\).
Now sum this identity from \(r=1\) to \(r=n\):
\(\displaystyle \sum_{r=1}^{n}\big((r+1)^4-r^4\big)=\sum_{r=1}^{n}(4r^3+6r^2+4r+1)\).
The left-hand side telescopes:
\((2^4-1^4)+(3^4-2^4)+\cdots+((n+1)^4-n^4)=(n+1)^4-1\).
Let \(S=\displaystyle \sum_{r=1}^{n}r^3\). Then
\((n+1)^4-1=4S+6\sum_{r=1}^{n}r^2+4\sum_{r=1}^{n}r+n\).
Using
\(\displaystyle \sum_{r=1}^{n}r=\frac{n(n+1)}{2}\), and \(\displaystyle \sum_{r=1}^{n}r^2=\frac{n(n+1)(2n+1)}{6}\),
we get
\((n+1)^4-1=4S+n(n+1)(2n+1)+2n(n+1)+n\).
Since \((n+1)^4-1=n^4+4n^3+6n^2+4n\), it follows that
\(4S=n^4+4n^3+6n^2+4n-n(n+1)(2n+1)-2n(n+1)-n\).
Simplifying the right-hand side gives
\(4S=n^4+2n^3+n^2=n^2(n+1)^2\).
Hence
\(\displaystyle \sum_{r=1}^{n}r^3=\frac{1}{4}n^2(n+1)^2\).
