User:Hdjur jcv/sandbox

Proof by Mathematical Induction

The equation obviously holds for n=1. For n+1 we have:

\begin{align} \sum_{k=1}^{n+1} k^3 &= \sum_{k=1}^n k^3 + (n + 1)^3 \\ \bigg(\sum_{k=1}^{n+1} k\bigg)^2 &= \bigg(\sum_{k=1}^n k + (n + 1)\bigg)^2 \\ &= \bigg(\sum_{k=1}^n k\bigg)^2 + 2(n+1)\sum_{k=1}^n k + (n + 1)^2 \\ &= \bigg(\sum_{k=1}^n k\bigg)^2 + 2(n+1)n(n+1)/2 + (n + 1)^2 \\ &= \bigg(\sum_{k=1}^n k\bigg)^2 + (n+1)n(n+1) + (n + 1)^2 \\ &= \bigg(\sum_{k=1}^n k\bigg)^2 + n(n+1)^2 + (n + 1)^2 \\ &= \bigg(\sum_{k=1}^n k\bigg)^2 + (n+1)^2 (n + 1) \\ &= \bigg(\sum_{k=1}^n k\bigg)^2 + (n+1)^3 \\ \end{align}$$ From there it can be seen that if the equation holds for n, it holds for n+1 too. In third step of the second derivation, well-known identity was used:

\begin{align} \sum_{k=1}^n k &= n(n+1)/2 \\ \end{align}$$