# Triangular Numbers

Alignments to Content Standards: F-LE.A

Below are pictures of the first three triangular numbers: In general, the $n^{\text{th}}$ triangular number is the total number of dots in $n$ columns where the columns have $1, 2, 3, \ldots, n$ dots.

1. The following picture relates the first three triangular numbers to areas of rectangles: Use this idea to calculate the seventh triangular number, $1 + 2 + 3 + 4 + 5+ 6+ 7$.
2. Calculate the hundredth triangular number, $1 + 2 + 3 + \cdots + 98 + 99 + 100$.
3. Find a formula for calculating the $n^{\text{th}}$ triangular number, $1 + 2 + \cdots + (n-1) + n.$

## IM Commentary

The goal of this task is to work on producing a quadratic equation from an arithmetic context. The teacher will have to closely monitor student work on part (c) as the question does not ask what type of ''formula'' is desired. The teacher might specify that it is a quadratic formula but this might be too leading and the groundwork for finding the quadratic formula has been laid in parts (a) and (b).

Part (b) has a famous history as a teacher reportedly assigned this problem to the famous mathematician Gauss as a young child. According to the story, Gauss solved the problem very quickly, in the manner described in this task. There are many accounts of this, one of which can be found here: http://nzmaths.co.nz/gauss-trick-staff-seminar.

The teacher may wish to remove part (a) altogether or perhaps simply ask students to calculate the sum $1 + 2 + \ldots + 7$. Some students may just find the sum by performing the addition while others may work with the geometric picture of the triangle. Those who calculate the sum directly will need to look for a different method when they work on (b) and (c). The teacher will need to watch students carefully and provide guidance as needed. One interesting fact which the students may discover is that the sum of consecutive triangular numbers is always a perfect square: this is seen in the pictures of part (a) if the top row of blue dots is removed. The picture shows that the converse is also true: every perfect square larger than 1 is the sum of consecutive triangular numbers. Students may be asked to show this as an additional part to the question.

## Solution

1. Following the pattern in the pictures, if we put two of the triangles with $1 + 2 + \ldots + 7$ dots together we get an 8 by 7 rectangle of dots. Since 8 $\times$ 7 = 56, this means that $$1 + 2 + 3 + 4 + 5 + 6 + 7 = \frac{56}{2} = 28.$$

2. For the sum of the first 100 numbers, the same method works. If we add a second identical sum to the first, we can group 1 and 100 to make 101. We can group 2 and 99 to make 101. We can similarly group all other numbers into pairs, ending with 100 and 1. In all we have 100 groups of 101. One sum of the first 100 whole numbers will be half of this so $$1 + 2 + \ldots + 99 + 100 = \frac{100 \times 101}{2} = 5050.$$

3. The same technique applies to the sum of the first $n$ whole numbers. We add another identical sum and then group $1$ and $n$, $2$ and $n-1$, and so on through $n$ and $1$. As above, we find $n$ total groups with $n+1$ in each group so $$1 + 2 + \ldots + n = \frac{n(n+1)}{2}.$$ Hence the sum of the first $n$ whole numbers is represented by the quadratic expression $\frac{n(n+1)}{2}$.