Next Three Terms
From the above triangular pattern, we can deduce a general statement which can represent the nth triangular number in terms of n. Butin order to do this, a table of n, Tn,1st difference and 2nd difference should be drawn:
From the above table, we can see that our variables are n and Tn. When we try to classify this pattern, we can see that it is not arithmetic. Arithmetic sequences require a common difference (d). Meaning that when we subtract
from
we should get the same value each time
This continues so that the value in between each increases by 1. The common difference is not equal we do not have an arithmetic sequence. The sequence is also not geometric as it would have a common ratio (r). If we were to have a geometric series when we divide
the value of r will equal.
So, in our case, this particular triangular pattern sequence is not considered in the above two categories. This sequence can be categorized in the ‘special sequences’ which consists of different series such as triangular numbers, square numbers, cube numbers, Fibonacci numbers and more. One of the reasons why this sequence is called triangular numbers is because if the sequence is physically drawn, the shape of the diagram would be a triangle with dots equally spaced out, hence the name ‘triangular numbers’.
So we started from the this point with the equation
Using the 1st term, n = 1, we multiplied this value to the equation
using n=1
This is not the value of the 1stterm, we expect it to be one. If we divide this figure by 2, we will get the expected first term.
Therefore, the equation for the triangular numbers is
When we test this for the 7th term, n=7
When we test this for the 8th term, n=8
We are now going to find out more about special numbers, but this time investigating stellar numbers. Stellar numbers are figurate numbers, based on the number of dots that can fit in a centered stellar or star shape. Below are the first four representations for a star with six vertices:
S1
S2
S3
S4
To get the next number in each series we have to add multiples of 12. Like the pattern above this pattern is also not arithmetic or geometric for the same reasons and it is evident that we add multiples of 12 each time. Example: 1+12 = 13, 13+(12×2) = 37, 13+(12×3) = 73.In this case the common number is 12.
When Sn= S2,
13 – 1 = 12 (which is 12x1)
When Sn= S3,
37 – 1 = 36 (which is 12x3)
When Sn= S4,
73 – 1 = 72 (which is 12x6)
With a common figure we can deduce that this equation can be work out with a similar approach to the first. As this shape has six sides we can use this as a multiplier.
6n(n+12)
If n = 1
6(13) = 78. This figure is too big proving the equation to be wrong.
6n(n+1)
6(2) = 12.
This figure is also too big. There is something missing from the equation to decrease the ending number.
However, if n the expression inside the bracket was (n-1), we could arrive at 0, then add 1 t get the value expected.
6n(n-1) +1
6(1)(1-1) + 1 = 1
For n= 2
6n(n-1) +1
6(2)(2-1) + 1 = 13
As we continue to test this pattern it works for all the other term numbers. Therefore the equation for the series of this pattern = 6n(n-1) +1.
Now we will test it for other p stellar shapes. When the value of p = 6, this was a constant in the first equation. To being, we will look at a 7 stellar shape, and test to see if we can replace the constant 6 in the previous equation with 7 first. Like for the 6 stellar shape we will add multiples of seven to each figure to get the total number of dots.
Substituting 7 into the formula to replace 6:
6n(n-1) +1
7n(n-1) + 1
When n = 1
7(0) = 1 = 1
n = 2
14(1) + 1 = 15
This was tested for all the other 5 terms and was found to be correct.
Next, we tried a 5 stellar shape.
81
Substituting 5 into the formula to replace 6:
6n(n-1) +1
5n(n-1) + 1
When n = 1
3(0) +1 = 1
n = 2
5(2)(2-1) + 1 = 11
This was tested for all the other 5 terms and was found to be correct.
Therefore the general statement is pn(n-1) + 1
To arrive at this formula, we noticed that p was the stellar star shape and when n was put into the formula, we would be able to achieve the values required. However as this is a stellar star shape, we cannot calculate the series for a stellar star shape that is less than three. Therefore p>=3 in order for the formula to work.
In conclusion, we can say that geometric shapes lead to special numbers.
by
meddakik (student)
