If we put building bricks into rows, beginning from \(1\) brick in the first row, \(2\) bricks in the second, and so on, the total number of bricks is the \(n\)-th triangle number \(\Delta_n\):
|\[\begin{array}{c} \fbox x \end{array}\]|\[\begin{array}{cc} \fbox x\\ \fbox x&\fbox x \end{array}\]|\[\begin{array}{ccc} \fbox x\\ \fbox x&\fbox x\\ \fbox x&\fbox x&\fbox x \end{array}\]|\[\begin{array}{cccc} \fbox x\\ \fbox x&\fbox x\\ \fbox x&\fbox x&\fbox x\\ \fbox x&\fbox x&\fbox x&\fbox x \end{array}\]|\[\begin{array}{ccccc} \fbox x\\ \fbox x&\fbox x\\ \fbox x&\fbox x&\fbox x\\ \fbox x&\fbox x&\fbox x&\fbox x\\ \fbox x&\fbox x&\fbox x&\fbox x&\fbox x \end{array}\]|\(\cdots\)|
| \(\Delta_1:=1\) | \(\Delta_2:=3\) | \(\Delta_3:=6\) | \(\Delta_4:=10\) | \(\Delta_5:=15\) | \(\cdots\) |
|---|---|---|---|---|---|
The general formula for the \(n\)-th triangle number is
\[\Delta_n=\sum_{k=1}^n n=1+2+\cdots+n=\frac {(n+1)n}{2},\]
which can easily be proven as a special case of the sum of arithmetic progression.
The sequence of triangle numbers begins with \(1\), \(3\), \(6\), \(10\), \(15\), \(21\), \(28\), \(36\), ...
Explanations: 1
