Triangular Numbers

Zaffy

Okay well a few days ago I was just bored in class and looking at triangular numbers.

Figured out that (n^2 + n)/2 = T when the number in the triangle number when n = number along, and T = the total amount.

So n = 1, T = 1
n = 2, T = 3
n = 3, T = 6
etc

http://milan.milanovic.org/math/english/triangular/trisqpent.gif

But what I was wondering was is there a way to figure out a formula to figure out the sum of all the previous triangular numbers. So say we were trying to figure out the first 3, would there be a way to add it up apart from
(1^2 + 1)/2 + (2^2 + 2)/2 + (3^2 + 3)/2
etc.

Any suggestions? I was just bored.

Fringe

We have 1/2(n^2 + n). Look at the two separately. Do you know how to sum the naturals and the naturals squared?

hivizman

If you think about the triangular numbers geometrically, they can be represented as two-dimensional arrays of points where there are n points in the bottom row, n-1 points in the row above this, each point being above and midway between two adjacent points in the bottom row, and so on until there is only one point in the top row.

The three-dimensional equivalent to this is a tetrahedron formed of one point at the top, three points in a triangle in the second layer, six points in a triangle in the third layer, and so on, with the points in each layer being above and equidistant from three adjacent points (in a triangle) in the layer beneath. By analogy, the number of points in such a tetrahedron can be called a tetrahedral number, and the total number of points in a tetrahedron whose lowest layer is a triangle whose side has n points is simply the sum of the first n triangular numbers.

As noted by Fringe, if you know the formula for the sum of the first n squares

n(n+1)(2n+1)/6

and the formula for the sum of the first n integers n(n+1)/2, you can represent the summation of the latter formula as equal to one half multiplied by (sum of squares from 1 to n PLUS sum of integers from 1 to n). Substituting the two summation formulae into this and rearranging gives a rather elegant formula for the nth tetrahedral number.

The sum of two consecutive triangular numbers must be a square. This can be seen geometrically, as you can divide a square consisting of nxn points into a triangle made up of 1 point from the first row down to n points in the last row, and then the remaining points in each row give you a triangle made up of n-1 points in the top row down to 1 point in the last but one row.

For example, 6^2 = 36 = 15+21. 15 is the 5th triangular number and 21 is the 6th triangular number.

Take the squares of integers from 1 to n. You can divide these up into triangular numbers as follows:

1 = 0+1
4 = 1+3
9 = 3+6
16 = 6+10
and so on down to
n^2 = (n-1)th triangular number + nth triangular number.

Add up all the squares and you have the sum equal to the sum of all triangular numbers from 1 to n-1 plus the sum of all triangular numbers from 1 to n. Hence, adding the nth triangular number to both sides and dividing both sides by two, we have one half multiplied by (sum of squares from 1 to n PLUS sum of integers from 1 to n) equals the nth tetrahedral number. This is a rather more roundabout, but perhaps more geometrical, way of getting Fringe's relationship.

Interestingly, the sum of the cubes of integers from 1 to n is equal to the square of the sum of the integers from 1 to n, that is, the square of the nth triangular number. For example, 1+8+27 = 36 = 6^2, and 1+8+27+64 =100 = 10^2.

Fremen

Nice way to see the sum-of-squares bit: