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

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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>.  
  
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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>.
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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>.
  
 
==See Also==
 
==See Also==

Revision as of 22:50, 22 November 2011

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 $x^{28}$?

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

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 $B$. For example, if I choose $x^2$ from $A$ , then there is exactly one power of $x$ in $B$ that I can choose; in this case, it would be $x^{24}$. 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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