Triangular numbers are defined as “the number of dots in an equilateral triangle uniformly filled with dots”. The sequence of triangular numbers are derived from all natural numbers and zero, if the following number is always added to the previous as shown below, a triangular number will always be the outcome:

1 = 1

2 + 1 = 3

3 + (2 + 1) = 6

4 + (1 + 2 + 3) = 10

5 + (1 + 2 + 3 + 4) = 15

Moreover, triangular numbers can be seen in other mathematical theories, such as Pascal’s triangle, as shown in the diagram below. The triangular numbers are found in the third diagonal, as highlighted in red.

The first diagrams to be considered show a triangular pattern of evenly spaced dots, and the number of dots within each diagram represents a triangular number.

Thereafter, the sequence was to be developed into the next three terms as shown below.

The information from the diagrams above is represented in the table below.

Term Number (n)| 1| 2| 3| 4| 5| 6| 7| 8|

Triangular Number (Tn)| 1| 3| 6| 10| 15| 21| 28| 36|

Establishing the following three terms in the sequence was done by simply drawing another horizontal row of dots to the previous equilateral and adding those dots to the previous count. However, following the method described earlier can also do this calculation, as shown in the illustration below. T11 = 1

T22 + 1 = 3

T33 + (2 + 1) = 6

T44 + (1 + 2 + 3) = 10

T55 + (1 + 2 + 3 + 4) = 15

T66 + (1 + 2 + 3 + 4 + 5) = 21

T77 + (1 + 2 + 3 + 4 + 5 + 6) = 28

T88 + (1 + 2 + 3 + 4 + 5 + 6 + 7) = 36

As seen in the diagram above, the second difference is the same between the terms, and the sequence is therefore quadratic. This means that the equation Tn = an2 + bn + c will be used when representing the data in a general formula. Since some of the values of Tn have already been established this makes it possible to work out the general formula. The first step is to substitute the established values into the three quadratic equations, as shown below:

When n = 1, Tn = 1

1 = a(1)2 + b(1) + c

1 = 1a + 1b + c

1 = a + b + c

When n = 2, Tn = 3

3 = a(2)2 + b(2) + c

3 = 4a + 2b + c

When n = 3, Tn = 6

6 = a(3)2 + b(3) + c

6 = 9a + 3b + c

Thereafter polysmlt on the graphic display calculator is used in order to retrieve the values of a, b, and c. The coefficients above are simply plugged into the calculator; this can be seen in the screenshots below

Hence the general formula is Tn = .5n2 + .5n, which can also be stated as Tn =

To be certain that the general formula in in fact correct, it can also be worked out as shown below.

The second difference of the sequence is divided by 2, since this is the value of a, thereafter the sequence is extended to T0, as the number highlighted in red is c. To figure out the value of b, the values have been substituted into the quadratic formula as shown below, after which the equation has been solved using simple algebra.

Tn = an2 + bn + c

Tn = ½n2 + bn + 0

T1 = (½)(1)2 + b(1)

1 = ½ + b

b = ½

∴ Tn =

As seen above, this method works out the same general formula, and this is tested below to assess the validity.

Tn = T3 = (32/2) + (3/2)T3 = (9/2) + (3/2)T3 = 12/2T3 = 6| Tn = T5 = (52/2) + (5/2)T5 = (25/2) + (5/2)T5 = 30/2T5 = 15| Tn = T12 = (122/2) + (12/2)T12 = (144/2) + (12/2)T12 = 156/2T12 = 78|

As seen above the general formula is indeed correct; however, this can also be determined by the method continued from page 2…

T99 + (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) = 45

T1010 + (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) = 55

T1111 + (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) = 66

T1212 + (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11) = 78

This furthermore proves the validity of the...

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