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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

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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