

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 12+34+56 = (12)+(34)+(56) = 1*3:
12+34+...+(N+1)N = 1*N/2, if N is even
If N is odd, things are similar, but not the same:
12+34+...+(N2)(N+1)+N = (12)+...+[(N2)(N1)] + N
12+34+...+(N2)(N+1)+N = 1 + 1 + ... + 1 + (N) = 1*(N1)/2 + N
12+34+...+(N2)(N+1)+N = [(N1)+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 byitself 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: (10010), (9911),
(9812), ..., (6050) 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 nontrivial
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
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Update: June 2012

