# triangular numbers

 $t_{n}=\sum_{i=1}^{n}i$

That is, the $n$th triangular number is simply the sum of the first $n$ natural numbers  . The first few triangular numbers are

 $1,3,6,10,15,21,28,\ldots$

The name triangular number comes from the fact that the summation defining $t_{n}$ can be visualized as the number of dots in

 $\begin{matrix}\bullet&&&&&&&\\ \bullet&\bullet&&&&&&\\ \bullet&\bullet&\bullet&&&&&\\ \bullet&\bullet&\bullet&\bullet&&&&\\ \bullet&\bullet&\bullet&\bullet&\bullet&&&\\ \bullet&\bullet&\bullet&\bullet&\bullet&\bullet&&\\ \vdots&&&&&\vdots&\ddots&\end{matrix}$

where the number of rows is equal to $n$.

The closed-form for the triangular numbers is

 $t(n)=\frac{n(n+1)}{2}$

Legend has it that a grammar-school-aged Gauss was told by his teacher to sum up all the numbers from 1 to 100. He reasoned that each number $i$ could be paired up with $101-i$, to form a sum of $101$, and if this was done $100$ times, it would result in twice the actual sum (since each number would get used twice due to the pairing). Hence, the sum would be

 $1+2+3+\cdots+100=\frac{100(101)}{2}$

The same line of reasoning works to give us the closed form for any $n$.

Another way to derive the closed form is to assume that the $n$th triangular number is less than or equal to the $n$th square (that is, each row is less than or equal to $n$, so the sum of all rows must be less than or equal to $n\cdot n$ or $n^{2}$), and then use the first few triangular numbers to solve the general 2nd degree polynomial $An^{2}+Bn+C$ for $A$, $B$, and $C$. This leads to $A=1/2$, $B=1/2$, and $C=0$, which is the same as the above formula   for $t(n)$.

Title triangular numbers TriangularNumbers 2013-03-22 12:16:08 2013-03-22 12:16:08 akrowne (2) akrowne (2) 6 akrowne (2) Definition msc 11A99 msc 40-00