**Problem**: (2006 AMC 12A - #24) The expression

$ (x+y+z)^{2006}+(x-y-z)^{2006} $

is simplified by expanding it and combining like terms. How many terms are in the simplified expression?

**Solution**: This was probably my favorite problem (or #22) on the AMC-12A, though it should've been more around #20.

At first we're like oh, should've memorized trinomial expansion! And then we're like maybe we can work with binomial expansion instead since we actually know that.

Then try setting $ w = y+z $.

We now have $ (x+w)^{2006}+(x-w)^{2006} $. It's easy to see that when $ w $ is taken to an odd power the terms from the first expansion and the second expansion will cancel each other out. Using the Binomial Theorem we have the remaining even terms to be

$ 2(x^{2006}+2006C2 \cdot x^{2004}w^2+\cdots+w^{2006}) $

Since each of these terms have a different power of $ x $, we just need to find the number of terms within the expansion of $ w = y+z $ to some power. Using the Binomial Theorem again, we find that $ w^n = (y+z)^n $ simply has $ n+1 $ terms, which should be pretty clear.

Then the total number of terms, by just using the even powers above and ignoring any constants and $x$'s since they don't affect the terms, is $ (0+1)+(2+1)+(4+1)+\cdots+(2006+1) $ from each of $ (y+z)^0, (y+z)^2, \ldots, (y+z)^{2006} $. This simplifies to

$ 1+3+5+\cdots+2007 $,

and a handy formula tells us that the sum of the first $ n $ positive odd integers is just $ n^2 $, so since $ 2007 $ is the $ 1004 $th positive odd integer we have

$ 1+3+5+\cdots+2007 = 1004^2 = 1008016$

different terms. QED.

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Comment: A fun problem, quick and slick to do on the actual test. It's a good lesson that multinomial expansions are no fun and usually the binomial expansion is good enough (or you could just extend the idea).

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Practice Problem: (1990 AIME - #2) Evaluate $ (52+6\sqrt{43})^{\frac{3}{2}}-(52+6\sqrt{43})^{\frac{3}{2}} $.

i dunno if it was a typo or not, but the problem was:

ReplyDelete(x+y+z)^2006 + (x-y-z)^2006

not 2005 for the second term. i'm unsure if that changes your proof- you are so above my level it's astonishing.

however, i wanted your input on how i did the problem:

to be fair, i did expand the expressions on my 89 (god love it) but theoretically i could have done it by hand-

i expanded those expressions out for different n values, where n is the exponent they are both raised to. when n=1, # terms = 1

n=2, #=4

n=3, #=4

n=4, #=9

n=5, #=9

at that point, i looked over the numbers and deduced a formula, that the number of terms of that expansion raised to the n equals

([n/2] + 1)^2

where [n/2] is the floor of n/2. plugging in 2006, one gets 1004^2

Anyway- any thoughts on this (basically) guess and check method? Is there any way I could have figured it out with some sort of solid math?

You run an awesome site, by the way.

Oops. You're right; that was a typo. I did everything assuming that was a 2006. The method you described "works" for this problem because the pattern does continue. With "solid math," you'd just see that if you replaced 2006 with 2n, it's just the sum of the first n+1 odd numbers (or, equivalently, (n+1)^2). The proof of this is just a generalization of what I have posted; it's a bit complicated to describe besides what I've already said, but if you work out the algebra by expanding it with the binomial theorem you should be able to get the result.

ReplyDelete