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Formula for the sum of the cubes from 1 to n

by RoRi

Prove that

    \[ \sum_{k=1}^n k^3 = \frac{n^4}{4} + \frac{n^3}{2} + \frac{n^2}{4}. \]

Proof. We follow a similar strategy to this one. From here, here, and here we have

    \[ \sum_{k=1}^n 1 = n, \qquad \sum_{k=1}^n k = \frac{n^2}{2} + \frac{n}{2}, \qquad \sum_{k=1}^n k^2 = \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}. \]

Then, since k^4 - (k-1)^4 = 4k^3 - 6k^2 + 4k -1 and by the telescoping property \sum_{k=1}^n (k^4 - (k-1)^4) = n^4, we have,

    \begin{align*} n^4 &= \sum_{k=1}^n (4k^3 - 6k^2 + 4k - 1) \\ &= 4 \sum_{k=1}^n k^3 - 6 \sum_{k=1}^n k^2 + 4 \sum_{k=1}^n k - \sum_{k=1}^n 1 \\ \implies \sum_{k=1}^n k^3 &= \frac{1}{4} \left( n^4 + 6 \sum_{k=1}^n k^2 - 4 \sum_{k=1}^n k + \sum_{k=1}^n 1\right) \\ &= \frac{n^4}{4} + \frac{3}{2}\left(\frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6} \right) - \left( \frac{n^2}{2} + \frac{n}{2} \right) + \frac{n}{4} \\ &= \frac{n^4}{4} + \frac{n^3}{2} + \frac{n^2}{4}. \qquad \blacksquare \end{align*}

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