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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>
  
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==Solution 1==  
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==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>.  
  

Latest revision as of 21:54, 9 January 2021

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 1 (easiest)

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 $224 \Rightarrow \boxed{C}$.

Solution 2

Let $P(x) = \left(1 + x + x^2 + \cdots + x^{14}\right)^2 = a_0 + a_1x + a_2x^2 + \cdots + a_{28}x^{28}$. Then the $x^{28}$ term from the product in question $\left(1 + x + x^2 + \cdots + x^{27}\right)(a_0 + a_1x + a_2x^2 + \cdots + a_{28}x^{28})$ is

$1a_{28}x^{28} + xa_{27}x^{27} + x^2a_{26}x^{26} + \cdots + x^{27}a_1x = \left(a_1 + a_2 + \cdots a_{28}\right)x^{28}$

So we are trying to find the sum of the coefficients of $P(x)$ minus $a_0$. Since the constant term $a_0$ in $P(x)$ (when expanded) is $1$, and the sum of the coefficients of $P(x)$ is $P(1)$, we find the answer to be $P(1) - a_0 = \left(1 + 1 + 1^2 + \cdots 1^{14}\right)^2 - 1 = 15^2 - 1 = 224 \Rightarrow \boxed{C}$.

Solution 3

We expand $(1 + x + x^2 + x^3 + \cdots + x^{14})^2$ to $(1 + x + x^2 + x^3 + \cdots + x^{14}) * (1 + x + x^2 + x^3 + \cdots + x^{14})$ and use FOIL to multiply. It expands out to:

$1 + x + x^2 + x^3 + x^4 + \cdots + x^{14} +$

$\qquad x + x^2 + x^3 + x^4 + \cdots + x^{14} + x^{15} +$

$\qquad \qquad x^2 + x^3 + x^4 + \cdots + x^{14} + x^{15} + x^{16} + \cdots$

It becomes apparent that

$(1 + x + x^2 + x^3 + \cdots + x^{14})^2 = 1 + 2x + 3x^2 + 4x^3 + \cdots + 15x^{14} + 14x^{15} + 13x^{16} + \cdots + x^{28}$.

Now we have to find the coefficient of $x^{28}$ in the product:

$(1 + 2x + 3x^2 + 4x^3 + \cdots + 15x^{14} + 14x^{15} + 13x^{16} + \cdots + x^{28}) \cdot (1 + x + x^2 + x^3 + \cdots + x^{27})$.

We quickly see that the we get $x^{28}$ terms from $x^{27} \cdot 2x$, $x^{26} \cdot 3x^2$, $x^{25} \cdot 4x^3$, ... $15x^{14} \cdot x^{14}$, ... $x^{28} \cdot 1$. The coefficient of $x^{28}$ is just the sum of the coefficients of all these terms. $1 + 2 + 3 + 4 + \cdots + 15 + 14 + 13 + \cdots + 4 + 3 + 2 = 224$, so the answer is $\boxed{C}$.

Solution 4

Rewrite the product as $\frac{(x^{28} - 1)(x^{15} - 1)(x^{15} - 1)}{(x - 1)^3}$. It is known that

\[\frac{1}{(1 - x)^n} = \binom{n - 1}{n - 1} + \binom{n}{n - 1}x + \binom{n + 1}{n - 1}x^2 + \binom{n + 2}{n - 1}x^3 + \cdots + \binom{n - 1 + k}{n - 1}x^k + \cdots .\]

Thus, our product becomes

\[-\left( x^{28} - 1 \right) \left( x^{15} - 1 \right) \left( x^{15} - 1 \right) \left( \binom{2}{2} + \binom{3}{2}x + \binom{4}{2}x^2 + \cdots \right).\]

\[= -\left( x^{28} - 1 \right) \left( x^{15} - 1 \right) \left( x^{15} - 1 \right) \left( 1 + 3x + 6x^2 + \cdots \right).\]

We determine the $x^{28}$ coefficient by doing casework on the first three terms in our product. We can obtain an $x^{28}$ term by choosing $x^{28}$ in the first term, $-1$ in the second and third terms, and $1$ in the fourth term. We can get two $x^{28}$ terms by choosing $x^{15}$ in either the second or third term, $-1$ in the first term, $-1$ in the second or third term from which $x^{15}$ has not been chosen, and the $\binom{15}{2}x^{13}$ in the fourth term. We get $\binom{15}{2} * 2 = 210$ $x^{28}$ terms this way. (We multiply by $2$ because the $x^{15}$ term could have been chosen from the second term or the third term). Lastly, we can get an $x^{28}$ term by choosing $-1$ in the first three terms and a $\binom{30}{2}x^{28}$ from the fourth term. We have a total of $1 + 210 - 435 = -224$ for the $x^{28}$ coefficient, but we recall that we have a negative sign in front of our product, so we obtain an answer of $224 \Rightarrow \boxed{(C)}$.

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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