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Difference between revisions of "2008 AMC 12A Problems/Problem 19"

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In the expansion of
 
In the expansion of
  
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<math>\left(1 + x + x^2 + \cdots + x^{27}\right)\left(1 + x + x^2 + \cdots + x^{14}\right)^2</math>,
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<cmath>\left(1 + x + x^2 + \cdots + x^{27}\right)\left(1 + x + x^2 + \cdots + x^{14}\right)^2,</cmath>
  
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what is the coefficient of <math>x^{28}</math>?
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what is the [[coefficient]] of <math>x^{28}</math>?
  
 
<math>\textbf{(A)}\ 195 \qquad \textbf{(B)}\ 196 \qquad \textbf{(C)}\ 224 \qquad \textbf{(D)}\ 378 \qquad \textbf{(E)}\ 405</math>
 
<math>\textbf{(A)}\ 195 \qquad \textbf{(B)}\ 196 \qquad \textbf{(C)}\ 224 \qquad \textbf{(D)}\ 378 \qquad \textbf{(E)}\ 405</math>
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==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}}
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[[Category:Introductory Algebra Problems]]

Revision as of 17:01, 20 February 2008

Problem

In the expansion of

\[\left(1 + x + x^2 + \cdots + x^{27}\right)\left(1 + x + x^2 + \cdots + x^{14}\right)^2,\]

what is the coefficient of $x^{28}$?

$\textbf{(A)}\ 195 \qquad \textbf{(B)}\ 196 \qquad \textbf{(C)}\ 224 \qquad \textbf{(D)}\ 378 \qquad \textbf{(E)}\ 405$

Solution

Let $A = \left(1 + x + x^2 + \cdots + x^{14}\right)$ and $B = \left(1 + x + x^2 + \cdots + x^{27}\right)$. We are expanding $A \cdot A \cdot B$.

Since there are $15$ terms in $A$, there are $15^2 = 225$ ways to choose one term from each $A$. The product of the selected terms is $x^n$ for some integer $n$ between $0$ and $28$ inclusive. For each $n \neq 0$, there is one and only one $x^{28 - n}$ in $B$. Since there is only one way to choose one term from each $A$ to get a product of $x^0$, there are $225 - 1 = 224$ ways to choose one term from each $A$ and one term from $B$ to get a product of $x^{28}$. Thus the coefficient of the $x^{28}$ term is $224 \Rightarrow C$.

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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