Squared Triangular Number - Proofs

Proofs

Charles Wheatstone (1854) gives a particularly simple derivation, by expanding each cube in the sum into a set of consecutive odd numbers:

\begin{align} \sum_{k=1}^n k^3 &= 1 + 8 + 27 + 64 + \cdots + n^3 \\ &= \underbrace{1}_{1^3} + \underbrace{3+5}_{2^3} + \underbrace{7 + 9 + 11}_{3^3} + \underbrace{13 + 15 + 17 + 19}_{4^3} + \cdots + \underbrace{\left(n^2-n+1\right) + \cdots + \left(n^2+n-1\right)}_{n^3} \\ &= \underbrace{\underbrace{\underbrace{\underbrace{1}_{1^2} + 3}_{2^2} + 5}_{3^2} + \cdots + \left(n^2 + n - 1\right)}_{\left( \frac{n^{2}+n}{2} \right)^{2}} \\ &= (1 + 2 + \cdots + n)^2 \\ &= \left(\sum_{k=1}^n k\right)^2. \end{align}

The sum of any set of consecutive odd numbers starting from 1 is a square, and the quantity that is squared is the count of odd numbers in the sum. The latter is easily seen to be a count of the form 1+2+3+4+...+n.

In the more recent mathematical literature, Stein (1971) uses the rectangle-counting interpretation of these numbers to form a geometric proof of the identity (see also Benjamin, Quinn & Wurtz 2006); he observes that it may also be proved easily (but uninformatively) by induction, and states that Toeplitz (1963) provides "an interesting old Arabic proof". Kanim (2004) provides a purely visual proof, Benjamin & Orrison (2002) provide two additional proofs, and Nelsen (1993) gives seven geometric proofs.

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