Diagonal Sum in Pascal's Triangle
Date: 04/02/2001 at 23:35:43 From: Gene Potter Subject: Pascal's Triangle Please explain how to find the sum of the reciprocals of the diagonals of Pascal's triangle.
Date: 04/03/2001 at 19:54:43 From: Doctor Schwa Subject: Re: Pascal's Triangle Hi Gene, so you mean, for instance, 1/1 + 1/3 + 1/6 + 1/10 + ... or in more compact math language, Sum [2/(n(n+1))] The nifty trick here is to "un-common-denominatorify" this fraction. That is, it now has a common denominator of n(n+1). What fractions A/n + B/(n+1) would add up to make this answer? (This method is known as partial fractions, and you can get help with it in the Dr. Math archives by searching for "partial fractions.") It turns out that 2/n - 2/(n+1) will do the trick. That is, you can rewrite 1/1 = 2/1 - 2/2 1/3 = 2/2 - 2/3 1/6 = 2/3 - 2/4 1/10 = 2/4 - 2/5 : Now the sum is easy to see: All the terms but 2/1 cancel, so the infinite sum is 2/1. The finite sum of the first n terms will be: 2/1 - 2/(n+1) What about the next diagonal? You could try a similar trick, but the algebra is bound to get messy. Well, not really: 6/(n(n+1)(n+2)) = 3/n - 6/(n+1) + 3/(n+2) If you have trouble doing that partial fractions trick, the archives will help. Here is a good starting point: What is a Partial Fraction? http://mathforum.org/dr.math/problems/ishmael1.8.99.html So again: 1/1 = 3/1 - 6/2 + 3/3 1/4 = 3/2 - 6/3 + 3/4 1/10 = 3/3 - 6/4 + 3/5 1/20 = 3/4 - 6/5 + 3/6 : By combining the terms with like denominators, you can see that this equals 3/1 - 3/2 in the long run; all the rest cancel, and the answer is 3/2. Now if you want a general formula for the k'th diagonal, I'd suggest doing one or two more by hand this way (it's not that hard), and then looking for a pattern and trying to prove it once you know what the answer is already... or write us back. This is an interesting problem. If you make some progress but get stuck somewhere along the line, please do let me know. I'm sure the answer is known, and I could just look it up somewhere along with a convenient trick for evaluating it, but it's more fun to try to figure it out for ourselves. Thanks for the fun problem, - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/
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