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:

\(\binom{x}{y}+\binom{x}{y+1}=\binom{x+1}{y+1}\)

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

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

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

2. The n^{th }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 6^{th} 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 S_{n }is the product of the terms in the n^{th }row.

**7) **

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

\(C_n=\binom{2n-3}{n-1}+\binom{2n-2}{n}-\binom{2n-3}{n}-\binom{2n-2}{n+1}\)

**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:

GYanggg Jun 26, 2018

#1**+3 **

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

TheXSquaredFactor Jun 26, 2018