2006 AMC 12A Problems/Problem 24

Problem

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?

$\mathrm{(A) \ } 6018\qquad \mathrm{(B) \ } 671,676\qquad \mathrm{(C) \ } 1,007,514$ $\mathrm{(D) \ } 1,008,016\qquad\mathrm{(E) \ } 2,015,028$

Solution

by the *Multinomial theorem, the expressions of the two are:

$\sum{\frac{2006!}{a!b!c!}x^ay^bz^c}$

and:

$\sum{\frac{2006!}{a!b!c!}x^a(-y)^b(-z)^c}$

respectively, where $a+b+c = 2006$ . Since the coefficients of like terms are the same in each expression, each like term either cancel one another out or the coefficient doubles. In each expansion there are:

${2006+2\choose 2} = 2015028$

terms without cancellation. For any term in the second expansion to be negative, the parity of the exponents of $y$ and $z$ must be opposite. Now we find a pattern:

if the exponent of $y$ is 1, the exponent of $z$ can be all even integers up to 2004, so 1003 terms.

if the exponent of $y$ is 3, the exponent of $z$ can go up to 2002, so 1002 terms.

$\vdots$

if the exponent of $y$ is 2005$, then $z$ can only be 0. So 1 term.

add them up we get $\frac{1003*1004}{2}$ terms. However, we can switch the exponents of $y$ and $z$ and these terms will still have a negative sign. So there are a total of $1003*1004$ negative terms.

Subtract this number from 2015028 we obtain $D. 1008016$ as our answer.

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2006 AMC 12A Problems/Problem 24

Problem

Solution

See also