Difference between revisions of "2008 AMC 12A Problems/Problem 19"
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<math>\mathrm{(A)}\ 195\qquad\mathrm{(B)}\ 196\qquad\mathrm{(C)}\ 224\qquad\mathrm{(D)}\ 378\qquad\mathrm{(E)}\ 405</math> | <math>\mathrm{(A)}\ 195\qquad\mathrm{(B)}\ 196\qquad\mathrm{(C)}\ 224\qquad\mathrm{(D)}\ 378\qquad\mathrm{(E)}\ 405</math> | ||
− | ==Solution 1== | + | ==Solution 1 (easiest)== |
Let <math>A = \left(1 + x + x^2 + \cdots + x^{14}\right)</math> and <math>B = \left(1 + x + x^2 + \cdots + x^{27}\right)</math>. We are expanding <math>A \cdot A \cdot B</math>. | Let <math>A = \left(1 + x + x^2 + \cdots + x^{14}\right)</math> and <math>B = \left(1 + x + x^2 + \cdots + x^{27}\right)</math>. We are expanding <math>A \cdot A \cdot B</math>. | ||
− | Since there are <math>15</math> terms in <math>A</math>, there are <math>15^2 = 225</math> ways to choose one term from each <math>A</math>. The product of the selected terms is <math>x^n</math> for some integer <math>n</math> between <math>0</math> and <math>28</math> inclusive. For each <math>n \neq 0</math>, there is one and only one <math>x^{28 - n}</math> in <math>B</math>. For example, if I choose <math>x^2</math> from <math>A</math> , then there is exactly one power of <math>x</math> in <math>B</math> that I can choose; in this case, it would be <math>x^{24}</math>. Since there is only one way to choose one term from each <math>A</math> to get a product of <math>x^0</math>, there are <math>225 - 1 = 224</math> ways to choose one term from each <math>A</math> and one term from <math>B</math> to get a product of <math>x^{28}</math>. Thus the coefficient of the <math>x^{28}</math> term is <math>224 \Rightarrow C</math>. | + | Since there are <math>15</math> terms in <math>A</math>, there are <math>15^2 = 225</math> ways to choose one term from each <math>A</math>. The product of the selected terms is <math>x^n</math> for some integer <math>n</math> between <math>0</math> and <math>28</math> inclusive. For each <math>n \neq 0</math>, there is one and only one <math>x^{28 - n}</math> in <math>B</math>. For example, if I choose <math>x^2</math> from <math>A</math> , then there is exactly one power of <math>x</math> in <math>B</math> that I can choose; in this case, it would be <math>x^{24}</math>. Since there is only one way to choose one term from each <math>A</math> to get a product of <math>x^0</math>, there are <math>225 - 1 = 224</math> ways to choose one term from each <math>A</math> and one term from <math>B</math> to get a product of <math>x^{28}</math>. Thus the coefficient of the <math>x^{28}</math> term is <math>224 \Rightarrow \boxed{C}</math>. |
== Solution 2 == | == Solution 2 == | ||
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= \left(1 + 1 + 1^2 + \cdots 1^{14}\right)^2 - 1 | = \left(1 + 1 + 1^2 + \cdots 1^{14}\right)^2 - 1 | ||
= 15^2 - 1 | = 15^2 - 1 | ||
− | = 224 \ | + | = 224 \Rightarrow \boxed{C}</math>. |
==Solution 3== | ==Solution 3== | ||
Line 55: | Line 55: | ||
==See Also== | ==See Also== | ||
+ | |||
{{AMC12 box|year=2008|ab=A|num-b=18|num-a=20}} | {{AMC12 box|year=2008|ab=A|num-b=18|num-a=20}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 22:54, 9 January 2021
Problem
In the expansion of
what is the coefficient of
?
Solution 1 (easiest)
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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