Difference between revisions of "2008 AMC 12A Problems/Problem 19"
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<cmath>= -\left( x^{28} - 1 \right) \left( x^{15} - 1 \right) \left( x^{15} - 1 \right) \left( 1 + 3x + 6x^2 + \cdots \right). </cmath> | <cmath>= -\left( x^{28} - 1 \right) \left( x^{15} - 1 \right) \left( x^{15} - 1 \right) \left( 1 + 3x + 6x^2 + \cdots \right). </cmath> | ||
− | We determine the <math>x^{28}</math> coefficient by doing casework on the first three terms in our product. We can obtain an <math>x^{28}</math> term by choosing <math>x^{28}</math> in the first term, <math>-1</math> in the second and third terms, and <math>1</math> in the fourth term. We can get two <math>x^{28}</math> terms by choosing <math>x^{15}</math> in either the second or third term, <math>-1</math> in the first term, <math>-1</math> in the second or third term from which <math>x^{15}</math> has not been chosen, and the <math>\binom{15}{2}x^{13}</math> in the fourth term. We get <math>\binom{15}{2} * 2 = 210</math> <math>x^{28}</math> terms this way. (We multiply by <math>2</math> because the <math>x^{15}</math> term could have been chosen from the second term or the third term). Lastly, we can get an <math>x^{28}</math> term by choosing <math>-1</math> in the first three terms and a <math>\binom{30}{2}x^{28}</math> from the fourth term. We have a total of <math>1 + 210 - 435 = -224</math> for the <math>x^{28}</math> coefficient, but we recall that we have a negative sign in front of our product, so we obtain an answer of <math>224</math> <math>\boxed{C)}</math>. | + | We determine the <math>x^{28}</math> coefficient by doing casework on the first three terms in our product. We can obtain an <math>x^{28}</math> term by choosing <math>x^{28}</math> in the first term, <math>-1</math> in the second and third terms, and <math>1</math> in the fourth term. We can get two <math>x^{28}</math> terms by choosing <math>x^{15}</math> in either the second or third term, <math>-1</math> in the first term, <math>-1</math> in the second or third term from which <math>x^{15}</math> has not been chosen, and the <math>\binom{15}{2}x^{13}</math> in the fourth term. We get <math>\binom{15}{2} * 2 = 210</math> <math>x^{28}</math> terms this way. (We multiply by <math>2</math> because the <math>x^{15}</math> term could have been chosen from the second term or the third term). Lastly, we can get an <math>x^{28}</math> term by choosing <math>-1</math> in the first three terms and a <math>\binom{30}{2}x^{28}</math> from the fourth term. We have a total of <math>1 + 210 - 435 = -224</math> for the <math>x^{28}</math> coefficient, but we recall that we have a negative sign in front of our product, so we obtain an answer of <math>224</math> <math>\boxed{(C)}</math>. |
==See Also== | ==See Also== |
Revision as of 12:31, 15 February 2016
Problem
In the expansion of
what is the coefficient of
?
Solution 1
Let and
. We are expanding
.
Since there are terms in
, there are
ways to choose one term from each
. The product of the selected terms is
for some integer
between
and
inclusive. For each
, there is one and only one
in
. For example, if I choose
from
, then there is exactly one power of
in
that I can choose; in this case, it would be
. Since there is only one way to choose one term from each
to get a product of
, there are
ways to choose one term from each
and one term from
to get a product of
. Thus the coefficient of the
term is
.
Solution 2
Let . Then the
term from the product in question
is
So we are trying to find the sum of the coefficients of minus
. Since the constant term
in
(when expanded) is
, and the sum of the coefficients of
is
, we find the answer to be
Solution 3
We expand to
and use FOIL to multiply. It expands out to:
It becomes apparent that
.
Now we have to find the coefficient of in the product:
.
We quickly see that the we get terms from
,
,
, ...
, ...
. The coefficient of
is just the sum of the coefficients of all these terms.
, so the answer is
.
Solution 4
Rewrite the product as . It is known that
Thus, our product becomes
We determine the coefficient by doing casework on the first three terms in our product. We can obtain an
term by choosing
in the first term,
in the second and third terms, and
in the fourth term. We can get two
terms by choosing
in either the second or third term,
in the first term,
in the second or third term from which
has not been chosen, and the
in the fourth term. We get
terms this way. (We multiply by
because the
term could have been chosen from the second term or the third term). Lastly, we can get an
term by choosing
in the first three terms and a
from the fourth term. We have a total of
for the
coefficient, but we recall that we have a negative sign in front of our product, so we obtain an answer of
.
See Also
2008 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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 |
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