Pascal's Triangle PatternDate: 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/ |
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