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

m (Reverted edits by Ryanyz10 (talk) to last revision by Nathan wailes)
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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 17:41, 29 June 2015

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

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 $224 \Rightarrow C$.

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 $P(1) - a_0 = \left(1 + 1 + 1^2 + \cdots 1^{14}\right)^2 - 1 = 15^2 - 1 = 224 \rightarrow C$


See Also

2008 AMC 12A (ProblemsAnswer KeyResources)
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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