= Square pyramidal number
{title2=$S_n$}
= Sum of squares of the first n positive integers
{synonym}
The <sum of squares> of the first $n$ <positive integers>: $S_n=\sum_{k=1}^n k^2=n(n+1)(2n+1)/6$. It counts a square-based stack with layers containing $1^2,2^2,\ldots,n^2$ objects. Differences $S_b-S_{a-1}$ give sums over consecutive squares.
The number of objects in a square-based pyramid with successive square layers of sizes $1^2,2^2,\ldots,n^2$. If $F(n)=n(n+1)(2n+1)/6$, direct algebra gives $F(n)-F(n-1)=n^2$ and $F(0)=0$. <Mathematical induction> therefore proves the displayed <summation> formula. Its leading term is $n^3/3$, so this <sum> grows cubically, even though its largest individual summand grows quadratically.
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