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A Power-Full Problem

At a mathematics conference recently, I picked up an interesting problems from one of the speakers (C.B., CWU).

Problem: Evaluate 1002- 992+982-972+962-952+...+22-12

The problem brought back memories of when I once asked students to solve the following: (A-X)(B-X)(C-X)(D-X)....(Y-X)(Z-X)!


Hint: What algebraic technique begs to be applied? Don't use a calculator.


Solution Commentary: I refuse to spoil it for you...You will know when you have the solution.

J.S. (Meridian, WA) writes: Believe it or not, I use this sort of thing to amuse myself while abusing myself on a treadmill. While doing so, recently, I thought of two ways to do this problem, although they are really the same and the second one is much quicker and shows clearly that my brain is too slow to be considered quick. I say that because the second method did not occur to me first.

Anyway, if one looks at this problem, it seems that by proper use of parentheses, one can change 100^2 - 99^2 + 98^2 - 97^2 + ... into (100^2 - 99^2) + (98^2 - 97^2) + ... and still mean the same thing.

The general form is then (n^2 - (n - 1)^2). Then my brain started to make things more difficult than necessary by looking at that as (n^2 - (n^2 - 2n + 1)) which, of course equals 2n-1. This is then an arithmetic series which appears much easier to deal with. Substituting into the formula n/2(a sub 1 + a sub 50) gives 50/2(199 + 3), 25(202), or 5050.

Having done this in my head while on a treadmill, I was feeling a bit haughty. Then, the next day, when on the treadmill again, I thought about the fact that it would have been much simpler to view (n^2 - (n - 1)^2) as (n + (n - 1))(n - (n-1)). In that case, the second factor is always 1, which immediately leaves us with (n + (n - 1))(1), or 2n - 1.

I'm sure you have gotten many answers similar to mine, but I suspect they found it easier to do than I did. Thanks for giving me something to do while I'm walking....