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For every nonnegative integer $n$,For example, $1^3+2^3+3^3+4^3=1+8+27+64=100=T_4^2$. A short proof uses the triangular number recurrence and the triangular number square-complement identity:Summing this difference of squares identity gives a telescoping sum; all intermediate squared triangular numbers cancel. Nicomachus describes a related grouping of the odd integers: take one, then the next two, then the next three, producing $1,3+5,7+9+11,\dots$, which are $1^3,2^3,3^3,\dots$. This historical grouping is recorded in Introduction to Arithmetic by Nicomachus (Hahn teacher edition) (Book II, Chapter 20, §5, printed p. 75).

A triangular number counts rows of $1,2,\dots ,n$ objects, with $n$ a nonnegative integer:The first values are $0,1,3,6,10,15,21,28,36,45,\dots$. The empty arrangement contributes zero, by the empty sum convention. Adding one more row gives the triangular number recurrence $T_n-T_{n-1}=n$ for $n\ge 1$. The row lengths form an arithmetic progression, while their running totals are triangular numbers.

Pair the rows with a reversed copy: $1+n=2+(n-1)=\cdots =n+1$. There are $n$ such pairs, so two copies contain $n(n+1)$ dots. Dividing by two proves the formula. This rectangle picture explains the factor one-half without requiring a formula for a long summation. For example, one hundred rows contain $T_{100}=5050$ dots.

The same count appears whenever every pair of different objects must be considered once. Five participants make ten possible handshakes; five sensors have ten different pairwise comparisons; a complete graph on five vertices has ten edges. These are the triangular number pair-counting identity, with one fewer row than objects. The distinction prevents a common error: $T_5=15$ counts all pairs from six objects, whereas five objects give $T_4=10$.

The triangular number square-complement identity says that two neighboring triangular numbers fill a square. Combining that identity with the one-row difference makes the difference of squares $T_n^2-T_{n-1}^2$ equal to $n^3$. It leads to Nicomachus's theorem: the sum of the first $n$ cubes of numbers is the square of a number $T_n^2$. A change in the way one groups objects exposes a relation between summation, multiplication and geometry.

This geometric view is already visible in Nicomachus's Introduction to Arithmetic by Nicomachus (Hahn teacher edition): Book II, Chapter 8 describes growing triangular rows, Chapter 12 combines successive triangular counts into squares, and Chapter 20 groups successive odd integers into cubes of numbers.^[ref] The linked edition uses Martin Luther D'Ooge's translation, revised with modern teaching material by Nathanael Hahn. Our empty-row index $T_0=0$ is a modern convention; the historical passage starts from unity and calls three the first actual triangle.

For a positive integer $n$, adding the row of length $n$ to the previous arrangement gives $T_n=T_{n-1}+n$. Thus the first differences of the triangular numbers are $1,2,3,\dots$. The triangular number recurrence provides incremental computation: after $T_6=21$, adding seven gives $T_7=28$.