Difference between revisions of "2008 AMC 12A Problems/Problem 19"
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= 224 \rightarrow C</math> | = 224 \rightarrow C</math> | ||
+ | ==Solution 3== | ||
+ | We expand <math>(1 + x + x^2 + x^3 + \cdots + x^{14})^2</math> to <math>(1 + x + x^2 + x^3 + \cdots + x^{14}) * (1 + x + x^2 + x^3 + \cdots + x^{14})</math> and use FOIL to multiply. It expands out to: | ||
+ | |||
+ | <math>1 + x + x^2 + x^3 + x^4 + \cdots + x^{14} + </math> | ||
+ | |||
+ | <math>\qquad x + x^2 + x^3 + x^4 + \cdots + x^{14} + x^{15} + </math> | ||
+ | |||
+ | <math>\qquad \qquad x^2 + x^3 + x^4 + \cdots + x^{14} + x^{15} + x^{16} + \cdots</math> | ||
+ | |||
+ | It becomes apparent that | ||
+ | |||
+ | <math>(1 + x + x^2 + x^3 + \cdots + x^{14})^2 = 1 + 2x + 3x^2 + 4x^3 + \cdots + 15x^{14} + 14x^{15} + 13x^{16} + \cdots + x^{28}</math>. | ||
+ | |||
+ | Now we have to find the coefficient of <math>x^{28}</math> in the product: | ||
+ | |||
+ | <math>(1 + 2x + 3x^2 + 4x^3 + \cdots + 15x^{14} + 14x^{15} + 13x^{16} + \cdots + x^{28}) * (1 + x + x^2 + x^3 + \cdots + x^{27})</math>. | ||
+ | |||
+ | We quickly see that the we get <math>x^{28}</math> terms from <math>x^{27} * 2x</math>, <math>x^{26} * 3x^2</math>, <math>x^{25} * 4x^3</math>, ... <math>15x^{14} * x^{14}</math>, ... <math>x^{28} * 1</math>. The coefficient of <math>x^{28}</math> is just the sum of the coefficients of all these terms. <math>1 + 2 + 3 + 4 + \cdots + 15 + 14 + 13 + \cdots + 4 + 3 + 2 = 224</math>, so the answer is <math>\boxed{C}</math>. | ||
==See Also== | ==See Also== |
Revision as of 12:02, 4 July 2015
Contents
[hide]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
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.