Hello Math Enthusiasts,
27.01.2020 (2nd update, PJ Selangor, Malaysia).
This post is for math enthusiasts - not for students preparing to sit for IB math or IGCSE Math 0580, 0606, 0607 or CIE Math 9709 or Math 9231 unless the students would like to know beyond their exam syllabuses.
It is about the general formula /equation, N = (n...n+(p-1))!/p!, to work-out:
1) the individual formula for the oblique/diagonal number series in Pascal/Yang Hui triangle
2) the value of any n-th term in the oblique/diagonal number series in the triangle.
The Pascal /Yang Hui triangle:
You see, currently, students of binomial theorem know that the horizontal series of numbers /binomial coefficients can be found by using binomial theorem or, to a limited extent, the Pascal /Yang Hui triangle.
What about the terms in the oblique/diagonal series of binomial coefficients?
As you can see, the binomial, say (x + y) or (a + b):
- to the power 1 yields an oblique number /binomial coefficient series called 'natural numbers': 1 + 2 + 3 + 4 + ... + n, where the n-th term = n;
- to the power 2 yields an oblique number /BC series called triangular number series of: 1 + 3 + 6 + 10 + 15 + ... + n, where the n-th term =n ( n+1)/2;
- to the power 3 yields a diagonal number /BC series called tetrahedral number series of: 1 + 4 + 10 + 20 + ... + n, where the n-th term = n(n+1)(n+2)/6;
- to the power 4 yields a diagonal number /BC series of: 1 + 5 + 15 + 35 + ... + n, where the n-th term = n(n+1)(n+2)(n+3)/24.
The n-th term formula for each oblique/diagonal bc series is currently analysed]and work-out individually - but I saw a pattern that led me to a general formula that can be used:
1) to work-out the individual formula for the n-th term of any oblique/diagonal number /BC series; and hence
2) to find the value of any n-th term in any oblique/diagonal BC series in the Pascal / Yang Hui triangle.
I have googled /searched far and wide and it seems that no one had found it before!. ..
The General Formula
N = (n...n+(p-1))!/p!
where,
1) n refers to the n-th term of the diagonal bc series generated by the binomial to the power p;
2) (n...n+(p-1))! refers to an 'up factorial" idea introduced n developed by me, where,
- an up factorial that starts with n and ends with n is:
(n...n)! = n
(n...n)! = n
- an up factorial that starts with n and ends with (n+1) is:
(n...n+1)! = n(n+1)
(n...n+1)! = n(n+1)
- an up factorial that starts with n and ends with (n+2) is:
(n...n+2)! = n(n+1)(n+2); and, so on.
(n...n+2)! = n(n+1)(n+2); and, so on.
3) p! refers to the conventional down factorial which starts with p and ends with 1. Thus,
p! = p(p-1)(p-2)x...x3x2x1
5! = 5x4x...x2x1
1! = 1
0! = 1
In the Pascal/Yang Hui triangle (please see above):
When the power p=1, a natural number series (1+2+3+4+5+...+n) is generated. And, using the general formula
N = (n...n+(p-1))!/p! = (n...n+(1-1))!/1! = (n...n)!/1! = n i.e. N = n
N = (n...n+(p-1))!/p! = (n...n+(1-1))!/1! = (n...n)!/1! = n i.e. N = n
When p=2, a triangular number series (1+3+6+10+15+21+...+n) is generated. And, using the general formula
N = (n...n+(p-1))!/p! = (n...n+1)!/2! = n(n+1)/(2x1) = n(n+1)/2 i.e. N = n(n+1)/2
N = (n...n+(p-1))!/p! = (n...n+1)!/2! = n(n+1)/(2x1) = n(n+1)/2 i.e. N = n(n+1)/2
When p=3, a tetrahedral number series (1+4+10+20+35+...+n) is generated. And, using the general formula
N = (n...n+(p-1))!/p! = (n...n+2)!/3! =n(n+1)(n+2)/6 i.e. N = n(n+1)(n+2)/6
N = (n...n+(p-1))!/p! = (n...n+2)!/3! =n(n+1)(n+2)/6 i.e. N = n(n+1)(n+2)/6
When p=4, a number series (1+5+15+35+70+...+n) is generated. And, using the general formula
N = (n...n+(p-1))!/p! = (n...n+3)!/4! = n(n+1)(n+2)(n+3)/24 i.e. N = n(n+1)(n+2)(n+3)/24
N = (n...n+(p-1))!/p! = (n...n+3)!/4! = n(n+1)(n+2)(n+3)/24 i.e. N = n(n+1)(n+2)(n+3)/24
And, so on...for p = 5, p = 6,...
Hence, the General Formula, N = (n...n+(p-1))!/p!, is capable of finding out:
1) the individual n-th term formula for any oblique/diagonal number or binomial coefficient series s in the Pascal /Yang Hui triangle - by just substituting the p value into the general formula; and,
2) the value of any n-th term in any oblique/diagonal series of binomial coefficients in the Pascal /Yang Hui triangle - by substituting the n-th value into the individual formula obtained by substituting p!
1) the individual n-th term formula for any oblique/diagonal number or binomial coefficient series s in the Pascal /Yang Hui triangle - by just substituting the p value into the general formula; and,
2) the value of any n-th term in any oblique/diagonal series of binomial coefficients in the Pascal /Yang Hui triangle - by substituting the n-th value into the individual formula obtained by substituting p!
You'll note that to generate the General Formula, I have to introduce the idea and notation of an up factorial. This idea of 'up factorial' can be used to introduce the idea of partial 'down factorial': for example, 9x8x7x6 can be expressed as (9...6)! So, the binomial coefficient (BC) which is usually expressed as: BC = p!/(n!(p-n))! can now be easily expressed as a 'partial down factorial / another partial down factorial'.
Hahaha, critical comments, anyone?
