Formula for Subtraction Series ```Name: Lilly Status: student Age: N/A Location: N/A Country: N/A Date: 11/21/2005 ``` Question: Is there a formula for subtracting numbers in a series? (like Gauss' formula for adding numbers in a sequence) Replies: Lilly, I do not know the formula right off, but it should be easy to derive. First, consider how Gauss derived his formula: 1+2+3+4+....+97+98+99+100 = (1+100)+(2+99)+(3+98)+(4+97)+...+(50+51) 1+2+...+99+100 = 50*101, (1+...+N)= N*(N+1)/2 Consider 1-2+3-4+5-6 = (1-2)+(3-4)+(5-6) = -1*3: 1-2+3-4+...+(N+1)-N = -1*N/2, if N is even If N is odd, things are similar, but not the same: 1-2+3-4+...+(N-2)-(N+1)+N = (1-2)+...+[(N-2)-(N-1)] + N 1-2+3-4+...+(N-2)-(N+1)+N = -1 + -1 + ... + -1 + (N) = -1*(N-1)/2 + N 1-2+3-4+...+(N-2)-(N+1)+N = [-(N-1)+2N]/2 = (N+1)/2, if N is odd. Knowing how the famous ones figured out their formulas and theories helps you to figure out your own formulas and theories. History of science is important. Dr. Ken Mellendorf Physics Instructor Illinois Central College Gauss's formula can be figured out, and so can your subtracting formula: add 1 + 2 .... + 98 + 99 (that is, add 1 to 99) pair the 1 with the 99, 1+99 = 100 pair the 2 with the 98, 2+98 = 100 pair the 49 with the 51 49+51 = 100 and the 50 is left by itself, so the sum is 49 times 100 (you did this 49 times above), plus the 50 left over, or 4950. so this is n(n+1)/2 = 99(100)/2 = 4950 So now do this with subtraction -1 -2 -3 -4 - 99 (that is, subtract 1, then 2, then 3, etc from zero) pair the -1 with the -99 to get -1 + -99 = -100 pair the -2 with the -98 to get -2 + -98 = -100 so the subtraction gives 49 times 100 (again, you did this 49 times) minus the by-itself 50, or -4950 so I think -n(n+1)/2 works, just stick a minus sign in front. Steve Ross For those who may not be familiar with Gauss's addition formula: I assume that you are referring to the story about Gauss in grade school where the students were asked to add the numbers in the sequence 1, 2, 3, 4, ..., 100 (busy work): Gauss, already showing his mathematical skill, thought "out of the box", realizing that if divided the sequence into two parts, the integers 1, 2, 3, ...50 and the integers 51, 52, 53, ..., 100 and reversed this second sequence to read: 100, 99, 98, ..., 52, 51, 50 and added the two "half sequences: 100+1, 99+2, 98+3, ..., 51+50 that each pair summed to: 101, 101, ..., 101 so that there were 50 pairs of sums, each pair summing to 101 so that the answer was simply: 50 x 101 = 5050. Now to your question: Once you realize logic that Gauss used it is possible to generate many variations. In fact, it would be a good project to find out how many different variations you could come up with: For example, here are two sequences involving subtraction: Consider the sequence: 100, 99, 98, ...,50 and -0, -1, -2, ..., -49, -50. If you pair the differences in the order: (100 - 50),( 99 - 49), (98 - 48), ..., (52 - 2), (51 - 1), (50 - 0) you have 51 pairs each difference being equal to 50. So the difference is 51 x 50 = 2550. Consider the sequences: -10, -11, -12, ..., -58, -59, -60 and -50, -51, -52, ..., -100 then reverse the second sequence: -100, -99, -98, ... , -50 and add pairs: (-100-10), (-99-11), (-98-12), ..., (-60-50) you have 51 pairs, each with a value of -110 so that the sum is: 51 x (-110) = -5610. There are very many such procedures (probably an infinite number) in which pairing of numbers make computation simple. Also there are variations that would be challenging. For example, suppose you divide the sequence of positive integers into triplets rather than pairs, quartets rather than pairs? Even more challenging suppose you consider the rational numbers: n / m where 'n' and 'm' are integers, and n < m. Can you find non-trivial sequences of integers 'n' and 'm' in which the sum of n / m is a constant. Or can one show that no such sequences exist? I do not know the answer to that question. Vince Calder Click here to return to the Mathematics Archives

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