Difference between revisions of "2006 AMC 12A Problems/Problem 24"
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<math>(x+P)^{2006}+(x-P)^{2006}</math>. | <math>(x+P)^{2006}+(x-P)^{2006}</math>. | ||
− | Expanding this using binomial theorem gives <math>x^n+ | + | Expanding this using binomial theorem gives <math>(x^n+Px^{n-1}+...+P^{n-1}x+P^n)+(x^n-Px^{n-1}+...-P^{n-1}x+P^n)</math>, where <math>n=2006</math> (we may omit the coefficients, as we are seeking for the number of terms, not the terms themselves). |
− | Simplifying gives: <math>2(x^n+x^{n-2} | + | Simplifying gives: <math>2(x^n+x^{n-2}P^2+...+x^2P^{n-2}+P^n)</math>. Note that two terms that come out of different powers of <math>P</math> cannot combine and simplify, as their exponent of <math>x</math> will differ. Therefore, we simply add the number of terms produced from each addend. By the Binomial Theorem, <math>P^k=(y+z)^k</math> will have <math>k+1</math> terms, so the answer is <math>1+3+5+...+2007=1004^2=1,008,016 \implies \boxed{D}</math>. |
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== See also == | == See also == |
Revision as of 21:29, 10 August 2015
Problem
The expression
is simplified by expanding it and combining like terms. How many terms are in the simplified expression?
Solution 1
By the Multinomial Theorem, the summands can be written as
and
respectively. 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:
terms without cancellation. For any term in the second expansion to be negative, the parity of the exponents of and must be opposite. Now we find a pattern:
if the exponent of is 1, the exponent of can be all even integers up to 2004, so there are 1003 terms.
if the exponent of is 3, the exponent of can go up to 2002, so there are 1002 terms.
if the exponent of is 2005, then can only be 0, so there is 1 term.
If we add them up, we get terms. However, we can switch the exponents of and and these terms will still have a negative sign. So there are a total of negative terms.
By subtracting this number from 2015028, we obtain or as our answer.
Solution 2
Alternatively, we can use a generating function to solve this problem. The goal is to find the generating function for the number of unique terms in the simplified expression (in terms of ). In other words, we want to find where the coefficient of equals the number of unique terms in .
First, we note that all unique terms in the expression have the form, , where and is some constant. Therefore, the generating function for the MAXIMUM number of unique terms possible in the simplified expression of is
Secondly, we note that a certain number of terms of the form, , do not appear in the simplified version of our expression because those terms cancel. Specifically, we observe that terms cancel when because every unique term is of the form:
for all possible .
Since the generating function for the maximum number of unique terms is already known, it is logical that we want to find the generating function for the number of terms that cancel, also in terms of . With some thought, we see that this desired generating function is the following:
Now, we want to subtract the latter from the former in order to get the generating function for the number of unique terms in , our initial goal:
which equals
The coefficient of of the above expression equals
Evaluating the expression, we get , as expected.
Solution 3
Define such that . Then the expression in the problem becomes: .
Expanding this using binomial theorem gives , where (we may omit the coefficients, as we are seeking for the number of terms, not the terms themselves).
Simplifying gives: . Note that two terms that come out of different powers of cannot combine and simplify, as their exponent of will differ. Therefore, we simply add the number of terms produced from each addend. By the Binomial Theorem, will have terms, so the answer is .
See also
2006 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.