Supposedly the following proves the sum of the first-$n$-squares formula given the sum of the first $n$ numbers formula, but I don’t understand it.

This question already has an answer here: Proving Nicomachus's theorem without induction 20 answers

Summing the first $n$ first powers of natural numbers: $$\sum_{k=1}^nk=\frac12n(n+1)$$ and there is a geometric proof involving two copies of a 2D representation of $(1+2+\cdots+n)$ that form a $n\times(n+1)$-rectangle. Similarly, $$\sum_{k=1}^nk^2=\frac16n(n+1)(2n+1)$$ has a geometric proof (scroll down just a bit til you see the wooden blocks) involving six copies of a 3D representation of $(1^2+2^2+\cdots+n^2)$ […]

In his gorgeous paper “How to compute $\sum \frac{1}{n^2}$ by solving triangles”, Mikael Passare offers this idea for proving $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$: Proof of equality of square and curved areas is based on another picture: Recapitulation of Passare’s proof using formulas is as follows: $$\sum_{n=1}^\infty \frac{1}{n^2} = \sum_{n=1}^\infty \int_0^\infty \frac{e^{-nx}}{n}\; dx\; = -\int_0^\infty \log(1-e^{-x})\; […]

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