The triangular number is a polygonal number - a number that can be represented by a regular geometric arrangement of equally spaced points. As the name suggests, triangular numbers can be visualized as a triangle of points. The triangular numbers can be found in the third diagonal of Pascal`s triangle, starting at row 3. The first triangular number is 1, the second is 3, the third is 6, the fourth is 10, and so on. Triangular numbers can be calculated by adding up consecutive numbers. Here are 5 different ways to develop (and prove) the formula for the sum of the first n natural numbers: T(n) = 1 + 2 + 3 + 4 + . + n Method 1: Using Differences The following table shows the sum of some natural numbers. n 0 1 2 3 4 5 Sn 0 1 3 6 10 15 Δ1 2 3 4 5 Δ2 1 1 1 1 Sn is the sum of the numbers to n. Because we find that the second difference Δ2 produces constant values, we assume the formula for the sum of the natural numbers is a quadratic, of the form an2+bn+c. Using our values, we substitute 0, 1, and 3 in the Equation: n Equation Equation Number 0 c=0 1 1 a+b=1 2 2 4a+2b=3 3 Note in Equations 2 and 3 that c=0. By subtracting twice Equation 2 from Equation 3, we get 2a=1, So a=1/2 Substituting the value for a in Equation 2, we find that b is also 1/2, So the sum of the first n natural numbers is Method 2: Using Gauss Let us write the sum of the natural numbers up to n in two ways as: Sn=1+2+3+...+(n-2)+(n-1)+n and Sn =n+(n-1)+(n-2)+...+3+2+1. If we add these two series we get: 2Sn=(n+1)+(n+1)+...+(n+1). There are n of these (n+1)'s so 2Sn=n(n+1) therefore Sn=(n(n+1))/2. Method 3: Using Summation Note that the sum of the squares of the first n natural numbers is the sum of the first (n+1) minus (n+1)2. Expanding the (k+1)th term: Expanding (n+1)2, and rounding up similar terms: Which gives us the sum of the first n natural numbers: Method 4: By Finding a General Term Note the difference between the sum of the first n natural numbers, and the sum to (n-1) is n: Sn – S(n-1) = n. Write down the known differences, hoping that a pattern appears and write Sn-k, and then write down the nth term, from which a formula can be extracted. A pattern becomes clear. In general, the kth term: Note that the "n" term is always kn, and the other term is a sum of the natural numbers. Make k=n, and so get the nth term, noting S0=0. The sum above is one n short of Sn: Substituting , → And finally, . Method 5: Using a Visual Model We can visualize the sum 1+2+3+...+n as a triangle of dots. Numbers which have such a pattern of dots are called Triangle (or triangular) numbers, written T(n), the sum of the integers from 1 to n : n 1 2 3 4 5 6 T(n) as a sum 1 1+2 1+2+3 1+2+3+4 1..5 1..6 T(n) as a triangle ... T(n)= 1 3 6 10 15 21 For the proof, we will count the number of dots in T(n) but, instead of summing the numbers 1, 2, 3, etc up to n we will find the total using only one multiplication and one division! To do this, we will fit two copies of a triangle of dots together, one red and an upside-down copy in green. E.g. T(4)=1+2+3+4 + = Notice that we get a rectangle which is has the same number of rows (4) but has one extra column (5). So the rectangle is 4 by 5. It therefore contains 4x5=20 balls, but we took two copies of T(4) to get this. So we must have 20/2 = 10 balls in T(4), which we can easily check. This visual proof applies to any size of triangle number. Here it is again on T(5): + = So T(5) is half of a rectangle of dots 5 tall and 6 wide, i.e. half of 30 dots, so T(5)=15. References: http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/runsums/triNbProof.html http://www.trans4mind.com/personal_development/mathematics/series/sumNatural Numbers.htm http://people.bath.ac.uk/sjb37/pascals-triangle-2.gif http://www.puzzles.com/puzzlesineducation/HandsOnPuzzles/TriangularNumbers/ TriangularNumbers.gif .