Ng Yu Kai (Sri KL) Scored A (88%) in IGCSE Physics 0625 in Winter 2019 Exams.

18.01.2020 (PJ):

Two days ago , Ng Yu Kai of Sri KL Int'l School - who took IGCSE Physics 0625 tuition from me - texted to say that he scored 88% (A) for the subject in the Oct/Nov 2019 exams.

His 1st physics tuition lesson with me was on 30/06/2019 in which he told me he had not been doing well and had been struggling to pass physics all these while.

We started with 'Electricity, Electro-magnetism and Electronics" - the chapters he was dealing with in school then; then 'Atomic Physics'. His score for these chapters were near perfect in the mock/final exams - however, the scores for the chapters not yet covered by my tuition were quite a disaster. Overall, his score for physics in the mock/final exams were only 50% (pls see his texted msg below)

So, I need to do what I got to do as time was running out - to brush him up on the areas not yet covered. And, the final score in the formal exams in Oct/Nov 2019 was: A (88%).

Heartiest congrats to him!

Don't forget to work doubly hard in foundation/A-level because I have seen students who did well in IGCSE-level but not so well in CIE or IB Diploma level mainly because of over-confidence - 1 month holidays means 1 month break from study/tuition totally. 10-day break shd be enough - when the going gets tough, the tough must get going!

Haha, here are his texted messages to me on the result-day (16.01.2020) and I will post his result-slip here when I receive it. :)

General Formula to Find n-th Term of Any Diagonal/Oblique Series of Binomial Coefficients in the Pascal/Yang Hui Triangle

Hello Folks,

Permit me to make a post on math in this physics blog.

27.01.2020 (2nd update, PJ Selangor, Malaysia).

It 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. Of course, students who would like to know more than what is specified in their exam syllabus are welcome!

It is about my general formula /equation,   N = (n...n+(p-1))!/p!, that can 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 from the triangle above, 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 called pentatope number series: 1 + 5 + 15 + 35 + ... + n, where the n-th term = n(n+1)(n+2)(n+3)/24;

- to the power 5 yields the 5-simplex number series: ....

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

- an up factorial that starts with n and ends with (n+1) is:
(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.

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

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

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 

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

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!

You'll note that to generate the General Formula, I have to introduce the idea and notation of an 'up factorial'.

Hahaha, critical comments, anyone?