Difference Triangles
Exploring Difference Triangles, the X-gon, and Pascals Triangle
The purpose of this article is to show how Pascals triangle can be obtained by using difference triangles.
Lets create a difference triangle where the base and first number of every difference is 1.
Here are recorded the results that are found in the base.
Use X as a set of numbers in a set n of differences,where X contains all integers >=1 and n contains all integers >=0.
The sum of the X-th number in the n-th difference and The X-th number in the (n+1)-th difference yields the (X+1)-th number in the n-th difference.
So the first triangle looks like this
1 Fourth Difference (1) 1 2 Third Difference (1,2) 1 2 4 Second Difference (1–3) 1 2 4 8 First Difference (1–4) 1 2 4 8 16 0-Difference = Base (1–5)
Notice the base pattern 2^n
Make Base (X) the first term of the (X-1)-th difference.
16 8 24 4 12 36 2 6 18 54 1 3 9 27 81
Notice the new base pattern 3^n
This will be repeated and diagramed as an X-gon (X sided figure)
X^n,1^n
X^4,1
2
X^3,1 4
6^n 2 8 2^n
1296 X^2,1 4 16
216 2 8
1080 36 X,1 4 24
180 6 2 12
900 30 1 6 36
50 5 3 18
750 25 4 9 54
125 20 12 27
625 100 16 36 81
5^n 500 80 48 108 3^n
400 64 144
320 192
256
4^n
The numbers in between the center and outermost points (radius) of the X-gon are constants raised to the power of increasing values of n. The outermost points (circumference) are increasing values of X raised to the power of constants.
Using the addition method of the difference triangles, how would we write (X+1)^n in terms of powers of X? in other words, referring to the X-gon above, how should the numbers in a radius be added to result in a number in the following radius?
Lets test for various values of n.
n=0
Since (X)^0 = (X+1)^0, The only term to be added when n=0 equals
1X^0
n=1
Looking at the X-gon we see that when n is raised by one power, the previous process used must be repeated, but with a higher value of n. Since the previous process yielded 1X^0, this result will yield 1X^1.
The sum of these two results is
1X^1+1X^0= (X+1)^1
n=2
Follow the same method as above. The previous method returned 1X^1+1X^0,
so the current method will be 1X^2+1X^1.
(1X^2+1X^1)+ (1X^1+1X^0) = 1X^2+2X^1+1X^0 = (X+1)^2
n=3
Repeat.
Previous [(1X^2+1X^1)+ (1X^1+1X^0)]
+Current [(1X^3+1X^2)+ (1X^2+1X^1)] =
1X^3+3X^2+3X^1+1X^0 = (X+1)^3
Lets look at the results for (x+1)^n where n ranges from 0 to 3.
(X+1)^0 = 1X^0 = 1 (X+1)^1 = 1X^1+1X^0 = 1 1 (X+1)^2 = 1X^2+2X^1+1X^0 = 1 2 1 (X+1)^3 = 1X^3+3X^2+3X^1+1X^0 = 1 3 3 1
This new triangle is Pascals Triangle.
It follows a counting method different from difference triangles.
The sum of the X-th number in the n-th difference and the (X+1)-th number in the n-th difference yields the (X+1)-th number in the (n-1)-th difference.
It would take a lot of adding if we were to use the difference triangles in the X-gon to
compute (X+1)^10. However, using the Pascals Triangle which we have derived from it, the task becomes much simpler. Lets expand Pascals Triangle.
(X+1)^0 1 (X+1)^1 1 1 (X+1)^2 1 2 1 (X+1)^3 1 3 3 1 (X+1)^4 1 4 6 4 1 (X+1)^5 1 5 10 10 5 1 (X+1)^6 1 6 15 20 15 6 1 (X+1)^7 1 7 21 35 35 21 7 1 (X+1)^8 1 8 28 56 70 56 28 8 1 (X+1)^9 1 9 36 84 126 126 84 36 9 1 (X+1)^10 1 10 45 120 210 252 210 20 45 10 1
The final line of the triangle tells us that
(X+1)^10 = 1X^10 + 10X^9 + 45X^8 + 120X^7 + 210X^6 + 252X^5 + 210X^4 + 120X^3 + 45X^2 + 10X^1 + 1X^0.