Use induction to prove that, for every positive integer \(n\), \[ 1^3+2^3+\cdots+n^3=(1+2+\cdots+n)^2. \]
Solution
- Use the fact that \(\sum_{i=1}^n i = \frac 1 2 n(n+1)\), so \[\begin{align*} \sum_{i=1}^{n+1} i^3 &= \sum_{i=1}^{n} i^3 + (n+1)^3 \\ &= \left( \sum_{i=1}^{n} i \right)^2 + (n+1)^3 \\ &= \left( \frac 1 2 n(n+1) \right)^2 + (n+1)^3 \\ &= \frac 1 4 n^2(n+1)^2 + (n+1)(n+1)^2 \\ &= \frac 1 4 (n+1)^2 (n^2 + 4(n+1)) \\ &= \frac 1 4 (n+1)^2(n+2)^2 \\ &= \left( \frac 1 2 (n+1)(n+2)\right)^2 \\ &= \left( \sum_{i=1}^{n+1} i \right)^2. \qed \end{align*}\]