Summing Consecutive Integers
Date: 08/30/98 at 06:05:17 From: Simon Subject: Arithmetic Sequences Dear Dr. Maths, Could you assist me with this problem? Express 1994 as a sum of consecutive positive integers and show that this is the only way to do it. I have worked out the answer by using guess and check methods (the answer being 497 + 498 + 499 + 500 = 1994), but I do not know how to prove it is the only way. I believe there is something to do with the formula for calculating the sum of consecutive numbers in an arithmetic sequence. Thanks, Simon
Date: 08/31/98 at 12:45:23 From: Doctor Peterson Subject: Re: Arithmetic Sequences Hi, Simon. You're definitely on the right track, and in the process you've observed the big problem with guess-and-check. It's often the fastest way to solve a problem, but you don't come away with any confidence that you've found the answer. The formula for the sum of N numbers starting with M; that is, M + (M+1) + (M+2) + ... + (M+N-1) is: (2M+N-1)*N ---------- 2 as you can easily see by lining up two copies of the sum and adding them: M + (M+1) + (M+2) + ... + (M+N-1) (M+N-1) + (M+N-2) + (M+N-3) + ... + M ----------------------------------------------- (2M+N-1) + (2M+N-1) + (2M+N-1) + ... + (2M+N-1) So to make 1994 the sum of a series like this, you have to factor 2*1994 into two numbers that could be written as N and (2M+N-1); that is, into two factors A and B, from which we can get N = A and M = (B-A+1)/2. If I factor 2*1994 = 3988, I get 2*2*997 (997 is prime), so the only pairs I can choose to be A and B are A B --- --- 1 3988 2 1994 4 997 997 4 1994 2 3988 1 See what M and N are for each case, and you'll have your proof. (You'll notice that if you allow "consecutive positive integers" to be just one integer (N = 1) there is another answer!) I hope this method is clear enough for you. There may be other ways, but this is the first way I found. - Doctor Peterson, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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