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I've just started teaching myself algebra from a high school text-book and I stumbled upon this problem :

How many terms does the simplified form of $(a+b+c)^{2006}+(a-b-c)^{2006}$ have ?

I found something on Wikipedia that appeared to be useful, the Multinomial Theorem, but It still led me to no result . Can somebody walk me through the solution ?

I'm much more interested in the thinking rather than the actual number.

jake
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2 Answers2

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Hint. You need to calculate how many different terms of form $a^nb^mc^k$ can arise from expanding the power. Next you need to figure out what terms do cancel each other (one from $(a + b + c)^{2006}$ and the other from $(a - b- c)^{2006}$) and how many are those.

Edit. That hint was not very helpful. So, actually we need to count how many possible nonegative integer values $m, n, k$ satisfy $$ m + n + k = 2006\\ (-1)^m(-1)^k = 1 \Leftrightarrow m + k \text{ is even} $$ Since $m + k + n = 2006$, $m+k$ is even iff $n$ is even, i.e. $n = 2l$. Finally, the problem is to count how many solutions does the following equation have $$ 2l + m + k = 2006, \qquad l,m,k \geq 0 $$ For a fixed $l$ that equation has $2007 - 2l$ solutions. So the total number of solutions will be $$ \sum_{l=0}^{1003} 2007 - 2l = 1004 \cdot 2007 - 2 \frac{1003 \cdot 1004}{2} = 1004^2 $$

uranix
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You don't need the multinomial theorem right away. What you need to notice is that $(a+b+c)=a + (b + c)$ and $(a-b-c)=a-(b+c)$. This means that when you do the simple binomial expansion, say $x = a$ and $y = b+c$, you lose a bunch of terms. To be precise, you lose all the terms that would have an odd power of $y$. So now you are left with expanding the remaining $1004$ terms (you can check this number). The remaining terms get doubled.

So now you have $\sum_{i=0}^{1003} 2 {2006 \choose 2i} x^{2006-2i}y^{2i}$. I'm not quite sure how to complete it from here but I'm sure some algebraic manipulation (with some wrestling) would so the trick!

Jas
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  • Also, have you studied principle of inclusion and exclusion yet? Perfectly fine if not but I wonder if that might help! – Jas May 07 '15 at 19:01
  • Not yet, but there's no problem if you need to use it in your proof, I'll start studying it in this context. I just want to see a solution, this problem has been tearing me apart for the last hours – jake May 07 '15 at 19:05
  • Well, you need to group same powers so just count the number of powers of $a$ you can have and the powers of $b, c$ you can have, given that. $b, c$ range from $0$ to $2006$ while $a$ ranges in the even powers, $0$ to $2006$, but some of these can't occur simultaneously, so you want to exclude them. Does that help? If not I could look more into it! – Jas May 07 '15 at 19:13