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

(New page: ==Problem== In the expansion of <math>\left(1 + x + x^2 + \cdots + x^{27}\right)\left(1 + x + x^2 + \cdots + x^{14}\right)^2</math>, what is the coefficient of <math>x^{28}</math>? <mat...)
 
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==See Also==
 
==See Also==
{{AMC12 box|year=2007|ab=A|num-b=18|num-a=20}}
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{{AMC12 box|year=2008|ab=A|num-b=18|num-a=20}}

Revision as of 13:54, 17 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 (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