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In Courant's Differential and Integral Calculus (Vol. I), he presents the following derivation of a formula for the sum of the squares of the first $n$ integers:

Link to Extract

However, when I substitute $v=0,1,2,...n$ and sum all of the resulting equations, I find that

$\displaystyle (1^3+\cdots +(n+1)^3)-(1^3+\cdots +n^3)=3(1^2+\cdots +n^2)+3(1+\cdots +n)+1$

and hence that

$\displaystyle (n+1)^3=3S_2+3S_1+1$

Which does not lead to the correct formula, since the $n$ is missing.

Question: where does the $n$ on the RHS of Courant's result [i.e. $(n+1)^3=3S_2+3S_1+n+1$] come from?

A. Sable
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1 Answers1

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You are forgetting to sum the $1$ as well. Allow me to show you. $$(v+1)^3-v^3=3v+3v+1$$ $$\sum_{v=0}^{n} (v+1)^3-v^3 = \sum_{v=0}^{n} 3v^2+3v+1$$ $$(n+1)^3=3\Bigg(\sum_{v=0}^{n}v^2\Bigg) +3\Bigg(\sum_{v=0}^{n}v\Bigg)+ \Bigg(\sum_{v=0}^{n}1\Bigg)$$ And so by substituting $S_1$ and $S_2$, we get $$(n+1)^3=3S_2+3S_1+n+1$$

Isaac Browne
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