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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Just in case someone doesn't know, Pascal's Triangle looks like this:

```
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
: : : : :
```

The top number is 1, and there are ones along both sides. Every number in the middle is the sum of the two numbers above it. I've put the first 5 rows, but it goes on forever.

Pascal's triangle can be built in the way I just described, and another well-known way of building it is by row, using the combinatorial formula. The first row is 0C0 = 1, the second row is 1C0 and 1C1, the third row is 2C0, 2C1 and 2C2, and so on.

However, there's also an interesting way of building it by 'slanted column'.

Consider the sequence of numbers that starts with 1, and every subsequent element is the average of all the previous ones.

So we start with 1. The second element is then the average of 1, which is 1.

The third is the average of 1 and 1, and this isn't going anywhere interesting.

The sequence goes 1, 1, 1, ...

Now consider a similar sequence, but this time every element after the first is twice the average of the previous ones.

Twice the average of 1 is 2, so we start with 1, 2.

Twice the average of 1 and 2 is 1.5*2 = 3.

Continue with this and you get 1, 2, 3, 4, 5, ...

Now do the same thing but making a sequence out of "thrice-averages".

This one is 1, 3, 6, 10, ...

Writing these sequences diagonally and putting them on top of each other, along with others made from higher average-multipliers, gets you Pascal's triangle.

I haven't worked out why it happens, but it's certainly cool.

Why did the vector cross the road?

It wanted to be normal.

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,320

you mean

1

1 1

1 1+1 1

1 1+2 2+1 1

1 1+3 3+3 3+1 1?

This is already discovered by Chinese mathematician YANG hui, or Hui Yang, depending on the sequence.

And it does has a proof

you can think of it in combination intuitions

*Last edited by George,Y (2008-12-26 18:29:16)*

**X'(y-Xβ)=0**

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

no, you're going along the rows of the triangle there

```
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
: : : : :
```

he means to consider the diagonals like:

1,1,1,1,1

1,2,3,4,5

1,3,6,10

1,4,10,20

then each diagonal can be represented as the average of previous elements in diagonal, multiplied by diagonal number

diag 1: 1, 1 = 1, 1 = (1+1)/2, 1 = (1+1+1)/3 etc

diag 2: 1, 2 = 1*2, 3 = (1+2)/2*2, 4 = (1+2+3)/3*2 etc

diag 3: 1, 3 = 1*3, 6 = (1+3)/2*3, 10 = (1+3+6)/3*3 etc

diag 4: 1, 4 = 1*4, 10 = (1+4)/2*4, 20 = (1+4+10)/3*4 etc

so imagine if you will, the triangle is a slanted matrix

```
(i,j) (0,0)
(1,0)(0,1)
(2,0)(1,1)(0,2)
(3,0)(2,1)(1,2)(0,3)
(4,0)(3,1)(2,2)(1,3)(0,4)
: : : : :
```

*Last edited by luca-deltodesco (2008-12-26 22:31:46)*

The Beginning Of All Things To End.

The End Of All Things To Come.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Just to clarify, I'm not claiming to have discovered anything in that post.

I heard about the "by diagonals" method somewhere else, but I hadn't seen it before so I thought it was worth sharing.

Why did the vector cross the road?

It wanted to be normal.

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