Pascal's Triangle Pattern

Date: 04/22/99 at 08:35:52
From: bonnie dieter
Subject: Pascal's triangle and patterns

It seems that multiplying each entry, beginning from the second one, 
in each row of Pascal's triangle by 1, 2, 3, 4, 5... in order and 
adding these products, will yield a pattern. For instance,

3rd row  1*3 + 2*3 +3*1 = 3*2^2
4th row 1*4 + 2*6 + 3*4 + 4*1 = 4*2^2
5th row 1*5 + 2*10 + 3*10 + 4*5 + 5*1 = 5*2^2

What is the pattern? Express in terms of n, where n stands for the nth 
row. Can you explain or prove/disprove the pattern?

Thanks so much. You are a wonderful service to us all.

Thanks,
Bonnie

Date: 04/24/99 at 15:32:03
From: Doctor Floor
Subject: Re: pascals triangle and patterns

Hi, Bonnie,

Thanks for your question!

I think your statement has to be corrected a little:

3rd row  1*3 + 2*3 +3*1 = 3*2^2
4th row 1*4 + 2*6 + 3*4 + 4*1 = 4*2^3 (instead of 4*2^2)
5th row 1*5 + 2*10 + 3*10 + 4*5 + 5*1 = 5*2^4 (instead of 5*2^2)

The pattern is that if you multiply the numbers of the nth row of 
Pascal's triangle by the numbers 0,1,2,...,n respectively, then the 
sum of the products equals n*2^(n-1).

So, if we take the fourth row as an example:

       1 4 6 4 1

then 0*1 + 1*4 + 2*6 + 3*4 + 4*1 = 32 = 4*2^3.

For a proof we don't need to express the numbers in the nth row of 
Pascal's triangle in other mathematical expressions. We do need the 
nth row to have a row sum of 2^n, and to be a sort of palindrome: it 
must be the same whether you read it forward or backward.

Let us consider the sixth row as an example:

    1 6 15 20 15 6 1

We know that 1+6+15+20+15+6+1 = 64 = 2^6.

We have to explain why 0*1+1*6+2*15+3*20+4*15+5*6+6*1 = 6 * 2^5 = 3 * 
2^6.

Take a careful look at what happens with the two 15's in this row. One 
of them is multiplied by 2 and one by 4, so we could just as well have 
multiplied them both by 3. The same goes with the two 1s and the two 
6s. So, instead of multiplying by 0,1,2,3,4,5,6 respectively, we could 
have multiplied all by their average: 3. This works because of the 
symmetry. And it is easy to see that this average is always n/2 in the 
nth row.

So in the nth row, we see that instead of multiplying by 0,1,2,3,...,n 
respectively, we can multiply all the numbers in the row by n/2. Thus 
the new row sum becomes n/2 * 2^n = n*2^(n-1).

I hope this helped.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/