Integer Triangle That Sums to Cube

Bùi Quang Tuån has posted on the CutTheKnotMath an interesting observation that the sum of all integers in the diagram below

integer triangle whose terms add up to a cube

equals $n^{3}.$ Let's see why this is so. For the sum $S(n)$ of all the terms in the triangle (that consists of $n$ rows), adding the rows, we have:

$\begin{align}\displaystyle S(n)&=\sum_{k=1}^{n}\sum_{i=k}^{2n-1-k}i\\ &=\sum_{k=1}^{n}\bigg(\sum_{i=1}^{2n-1}i-(k-1)2n\bigg)\\ &=\sum_{k=1}^{n}\sum_{i=1}^{2n-1}i-2n\sum_{k=1}^{n}(k-1)\\ &=n\frac{(2n-1)2n}{2}-2n\frac{(n-1)n}{2}\\ &=n^{2}(2n-1)-n^{2}(n-1)\\ &=n^{3}. \end{align}$

We may also sum up columns first. The easiest way of doing that is by noticing that the columns have each consists of the same digit and the teram in two columns at the same distance (count) from the ends add up nicely: the first and the last columns give $1 + (2n-1)=2n.$ The second and the penultimate columns give $2[2+(2n-2)]=2\cdot 2n.$ For the third column and the third from the end we have $3[3+(2n-3)]=3\cdot 2n.$ And so on. The $n-$th column comes out unpaired with the sum of terms $n^{2}.$ Thus we have

$\begin{align}\displaystyle S(n)&=2n\sum_{k=1}^{n-1}k+n^{2}\\ &=2n\frac{(n-1)n}{2}+n^{2}\\ &=(n-1)n^{2}+n^{2}\\ &=n^{3}. \end{align}$

John Arioni has observed that, in the integer triangle, the middle (i.e., the logest column) has entries all equal to, say, $n$ whereas the entries to the left can be seen as differences $n-k$ and paired with the entries $n+k$ to the right of the middle column. If the pieces $\pm k$ are cancelled out, then the whole triangle appears to consist of entries, each equal to $n$. There are $2n-1$ of them in the first row, $2n-3$ in the second, etc., and just $1$ in the last, giving the sum:

$\displaystyle S(n)=n\sum_{k=1}^{n}(2k-1)=n\cdot n^{2}=n^{3}.$

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