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Recently, someone asked about the secrets of the Pascal's Triangle. My answer was not complete, since it didn't contain all the neat tricks and gimmicks. I have summed it all up here. This is the basic Pascal's Triangle:



1) Just looking at the numbers, we see that it can be represented by combinations. 


First row: \(\binom{0}{0}\)

Second row: \(\binom{1}{0}, \binom{1}{1}\)

Third row: \(\binom{2}{0}, \binom{2}{1}, \binom{2}{2}\)


This leads to our first identity, since each number is the sum of the two numbers above it, we get: 




2) Looking at each row, we get two conclusions: 


1. The sum of the numbers in the nth row is \(2^{n-1}\).


This gives us Pascal's Corollary 8: \( 2^n−1=2^{n−1}+2^{n−2}+...+1.\)


2. The nth row is \(11^{n-1}\)


\(1)\ 1=11^0\\ 2)\ 11=11^1\\ 3)\ 121=11^2\\ 4)\ 1331=11^3\\ 5)\ 14641=11^4\\ \)

For 6th row, you carry the additional digits up to their appropriate place, \(1\ \  \ 5\ \  \ 10\ \ \ 10\  \ \ 5\ \ \  1\) becomes \(1, 5+1, 0+1, 0 , 5, 1=161051=11^5\)   


3) Next, we look at each diagonal row. The first diagonal from the top, going down, is all ones. 


First row: only ones: 1, 1, 1, ...

Second row: counting numbers, 1, 2, 3, ...

Third row: triangular numbers, 1, 3, 6, ...

Fourth row: tetrahedral numbers, 1, 4, 10, ...


4) According to Pascal, in every arithmetical triangle each cell is equal to the sum of all the cells of the preceding row from its column to the first, inclusive (Corollary 2). Also, in every arithmetic triangle, each cell diminished by unity is equal to the sum of all those which are included between its perpendicular rank and its parallel rank, exclusively (Corollary 4).


Identity 1: \(C^{n+1}_{m+1}=C^n_m+C^{n-1}_m+\cdots+C^0_m\)


Identity 2: \(C^{n+1}_m-1=\sum C^k_j\)


5) We could detect the Fibonacci numbers if we cut the triangle in diagonals and take the sum of the diagonals: 



6) The fundamental constant e hidden in the Pascal Triangle; this by taking products - instead of sums - of all elements in a row:


\(\displaystyle\lim_{n \rightarrow \infty}\frac{S_{n-1}S_{n+1}}{s^2_n}=e\), where Sis the product of the terms in the nth row.





The nth Catalan number \(C_n=\frac{1}{n+1}\binom{2n}{n}\)could be discovered in the following combination:




8) Squares


Identity: \(C^{n+2}_3-C^n_3=n^2\\ C^{n+3}_4−C^{n+2}_4−C^{n+1}_4+C^n_4=n^2.\)


Also: \(\displaystyle\sum^n_{k=0}(C^n_k)^2=C^{2n}_n\)


9) Cubes



Identity: \(\displaystyle n^{3}=\bigg[C^{n+1}_{2}\cdot C^{n-1}_{1}\cdot C^{n}_{0}\bigg] + \bigg[C^{n+1}_{1}\cdot C^{n}_{2}\cdot C^{n-1}_{0}\bigg] + C^{n}_{1}\)


10) pi


Identity: \(\pi=3+\frac23(\frac1{C^4_3}-\frac1{C^6_3}+\frac1{C^8_3}-\cdots).\)


11) Fractals


If you shade in certain numbers, you get beautiful fractals:


 Jun 26, 2018
edited by GYanggg  Jun 28, 2018

It is amazing how all these hidden patterns can be extracted from a simple sequence of numbers formed in a triangular shape.  

 Jun 26, 2018



I totally agree with you, that's why I put together this list of all the patterns hidden inside the Pascal's  triange. 


I hope you enjoyed.

GYanggg  Jun 26, 2018

I enjoyed it and I enjoyed it well.

TheXSquaredFactor  Jun 26, 2018

Thanks Gavin,


Thanks for posting this great examination of Pascal's Triangle. 

I have posed a link to this in our Reference thread (it is in the sticky notes) so that people should be able to find it well into the future. 



 Jun 26, 2018

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