Everyone knows what squares are. The product of a number $$n$$ with itself is a square, so called because you can make an $$n \times n$$ array from $$n^2$$ objects. Few people, however, know about triangle numbers. Triangle numbers result from arranging objects to form a triangle.

Thus,

$$\begin {eqnarray} T_1 &=& 1 \\ T_2 &=& 3 \\ T_3 &=& 6 \\ T_4 &=& 10 \\ &\vdots& \end {eqnarray}$$

It should also be apparent that each $$T_n$$ is the sum of the first $$n$$ whole numbers. So,

$$T_n = 1 + 2 + 3 + \dots + n$$

Now suppose $$n$$ is even. Then we can rearrange the terms of the sum by pairing the first term with the last, second term with the second last, and so on to get,

$$n + 1 + [(n - 1) + 2] + [(n - 2) + 3] + \dots + [(\frac n 2 + 1) + \frac n 2 ]$$

The result is $$\frac n 2$$ terms all equal to $$(n + 1)$$. Thus,

$$T_{n_{even}} = {n(n + 1) \over 2}$$

This result can also be proved by mathematical induction. Similarly, if $$n$$ is odd, we get $$(n - 1) \over 2$$ terms all equal to $$(n + 1)$$ plus an additional term, $$(n + 1) \over 2$$. This also reduces to,

$$T_{n_{odd}} = {n(n + 1) \over 2}$$

Triangle numbers have a number of interesting properties. For example, the sum of two adjacent triangle numbers is always a perfect square. You can easily see this from the illustration.

You can also prove it by considering:

$$\begin {eqnarray} T_{n} + T_{n + 1} &=& {{n(n + 1)} \over 2} + {{(n + 1)(n + 2)} \over 2} \\ &=& {{n^2 + n + n^2 + 3n + 2} \over 2} \\ &=& {{2n^2 + 4n + 2} \over 2} \\ &=& n^2 + 2n + 1 \\ &=& (n + 1)^2 \end {eqnarray}$$

Note that if you sum the two sequences for $$T_{n - 1}$$ and $$T_{n}$$, you get a sum of a sequence of odd numbers. This shows that square numbers can be expressed as a sum of a sequence of odd numbers.

\begin {align} T_{n} &= &1 + &2 + 3 + 4 + \dots + n \\ T_{n - 1} &= & &1 + 2 + 3 + \dots + (n - 1) \\ T_{n} + T_{n - 1} &= &1 + &3 + 5 + 7 + \dots + (2n - 1) \\ n^2 &= &1 + &3 + 5 + 7 + \dots + (2n - 1) \end {align}