Subtracting Finite Sums of IntegersDate: 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/ |
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