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Subtracting Finite Sums of Integers


Date: 08/03/98 at 06:41:18
From: Sarah Wilson
Subject: Algebra

Let n = 1 + 3 + 5 + 7 + ... + 999
Let m = 2 + 4 + 6 + 8 + ... + 1000

Then m-n equals:

   A) 500
   B) 1000
   C) 499
   D) 501

I can't find a shortcut to work out the answer. The only way I can do 
this is with the calculator, going 1 + 3 + ... and so on. Can you 
please explain the answer and how you got it?

Thank you.


Date: 08/03/98 at 13:17:54
From: Doctor Rick
Subject: Re: Algebra

Hi, Sarah.

Let's look for a shortcut here. What if I write it this way:

     m         2 + 4 + 6 + 8 + ... + 1000
   - n  ==>  -(1 + 3 + 5 + 7 + ... +  999)
    --         ---------------------------
            =  1 + 1 + 1 + 1 + ... +    1

What I've really done is to use the commutative and associative 
properties of addition to rearrange the terms. If the sums were 
shorter, n = 1 + 3 + 5 and m = 2 + 4 + 6, you'd have:

   m - n = (2+4+6) - (1+3+5)
         = 2 + 4 + 6 + (-1 + -3 + -5)
         = 2 - 1 + 4 - 3 + 6 - 5
         = (2-1) + (4-3) + (6-5)
         = 1 + 1 + 1
         = 3

If this convinces you of what I wrote above, then all you have to do is 
find out how many 1's there are in that last sum. How many terms are 
there in m? (You should convince yourself that there are the same 
number of terms in n.)

- Doctor Rick, The Math Forum
Check out our web site! http://mathforum.org/dr.math/   
    
Associated Topics:
High School Sequences, Series

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