Difference between revisions of "Mock AIME 1 2006-2007 Problems/Problem 13"

Revision as of 11:26, 30 January 2009

Problem

Let $a_{n}$ , $b_{n}$ , and $c_{n}$ be geometric sequences with different common ratios and let $a_{n}+b_{n}+c_{n}=d_{n}$ for all integers $n$ . If $d_{1}=1$ , $d_{2}=2$ , $d_{3}=3$ , $d_{4}=-7$ , $d_{5}=13$ , and $d_{6}=-16$ , find $d_{7}$ .

Solution

The problem tempts us to simply introduce six variables: the first term and the ratio for each of the three geometric sequences. The problem statement then gives us six independent (non-linear) equations in these variables, which does uniquely determine these variables - but there is no obvious way of computing them. We will show a different solution.

Insight

Consider the polynomial $P(x)=(x-\alpha)(x-\beta)(x-\gamma)$ with $\alpha$ , $\beta$ and $\gamma$ distinct.

Let $\alpha+\beta+\gamma=p$ , $-(\alpha\beta+\alpha\gamma+\beta\gamma)=q$ , and $\alpha\beta\gamma=r$ . Then we can write $P(x)=x^3 - px^2 - qx - r$ .

Now consider the recurrence $\forall n: d_n = pd_{n-1} + qd_{n-2} + rd_{n-3}$ . ( $P$ is called the characteristic polynomial of this recurrence.)

We can easily verify that each of the three sequences $d_n=\alpha^n$ , $d_n=\beta^n$ and $d_n=\gamma^n$ satisfies the recurrence. Moreover, we know the following facts:

If a sequence $(x_n)$ satisfies the recurrence, then for all $s$ the sequence $(sx_n)$ does, too.
If sequences $(x_n)$ and $(y_n)$ satisfy the recurrence, then the sequence $(x_n+y_n)$ does, too.

Thus each linear combination of the three sequences we have satisfies the recurrence. In other words, each sequence of the form $d_n = s\alpha^n + t\beta^n + u\gamma^n$ solves the recurrence. And vice versa, each solution of the recurrence can be written in this form, because given the values $d_0$ , $d_1$ and $d_2$ one can always uniquely determine the coefficients $s,t,u$ .

Solving the task

We just proved the following observation: If we have three geometric sequences with distinct ratios $\alpha$ , $\beta$ , and $\gamma$ , their sum always satisfies the recurrence $\forall n: d_n = pd_{n-1} + qd_{n-2} + rd_{n-3}$ for the $p,q,r$ defined above, and some $d_0,d_1,d_2$ determined by the initial values of these sequences.

In our situation, we do not need to know the ratios $\alpha$ , \beta $, and$ \gamma $. All we need to compute$ d_7 $are the coefficients$ p $,$ q $, and$ r$.

But this is easy. We know that: <cmath> \begin{align*} d_4 &= pd_3 + qd_2 + rd_1 \\ d_5 &= pd_4 + qd_3 + rd_2 \\ d_6 &= pd_5 + qd_4 + rd_3 \end{align*} </cmath>

This is a set of three linear equations. In our case, it has a unique solution$ (Error compiling LaTeX. Unknown error_msg)(p,q,r)=(-2,-1,1) $, hence$ d_7 = -2d_6 - d_5 + d_4 = \boxed{012}$.

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Mock AIME 1 2006-2007

@@ Line 2: / Line 2: @@
 Let <math>a_{n}</math>, <math>b_{n}</math>, and <math>c_{n}</math> be geometric sequences with different common ratios and let <math>a_{n}+b_{n}+c_{n}=d_{n}</math> for all integers <math>n</math>. If <math>d_{1}=1</math>, <math>d_{2}=2</math>, <math>d_{3}=3</math>, <math>d_{4}=-7</math>, <math>d_{5}=13</math>, and <math>d_{6}=-16</math>, find <math>d_{7}</math>.
 ==Solution==
-{{solution}}
+The problem tempts us to simply introduce six variables: the first term and the ratio for each of the three geometric sequences. The problem statement then gives us six independent (non-linear) equations in these variables, which does uniquely determine these variables - but there is no obvious way of computing them. We will show a different solution.
+=== Insight ===
+Consider the polynomial <math>P(x)=(x-\alpha)(x-\beta)(x-\gamma)</math> with <math>\alpha</math>, <math>\beta</math> and <math>\gamma</math> distinct.
+Let <math>\alpha+\beta+\gamma=p</math>, <math>-(\alpha\beta+\alpha\gamma+\beta\gamma)=q</math>, and <math>\alpha\beta\gamma=r</math>. Then we can write <math>P(x)=x^3 - px^2 - qx - r</math>.
+Now consider the recurrence <math>\forall n: d_n = pd_{n-1} + qd_{n-2} + rd_{n-3}</math>. (<math>P</math> is called the ''characteristic polynomial'' of this recurrence.)
+We can easily verify that each of the three sequences <math>d_n=\alpha^n</math>, <math>d_n=\beta^n</math> and <math>d_n=\gamma^n</math> satisfies the recurrence. Moreover, we know the following facts:
+* If a sequence <math>(x_n)</math> satisfies the recurrence, then for all <math>s</math> the sequence <math>(sx_n)</math> does, too.
+* If sequences <math>(x_n)</math> and <math>(y_n)</math> satisfy the recurrence, then the sequence <math>(x_n+y_n)</math> does, too.
+Thus each linear combination of the three sequences we have satisfies the recurrence.
+In other words, each sequence of the form <math>d_n = s\alpha^n + t\beta^n + u\gamma^n</math> solves the recurrence. And vice versa, each solution of the recurrence can be written in this form, because given the values <math>d_0</math>, <math>d_1</math> and <math>d_2</math> one can always uniquely determine the coefficients <math>s,t,u</math>.
+=== Solving the task ===
+We just proved the following observation: If we have three geometric sequences with distinct ratios <math>\alpha</math>, <math>\beta</math>, and <math>\gamma</math>, their sum always satisfies the recurrence <math>\forall n: d_n = pd_{n-1} + qd_{n-2} + rd_{n-3}</math> for the <math>p,q,r</math> defined above, and some <math>d_0,d_1,d_2</math> determined by the initial values of these sequences.
+In our situation, we do not need to know the ratios <math>\alpha</math>, \beta<math>, and </math>\gamma<math>. All we need to compute </math>d_7<math> are the coefficients </math>p<math>, </math>q<math>, and </math>r<math>.
+But this is easy. We know that:
+<cmath>
+\begin{align*}
+d_4 &= pd_3 + qd_2 + rd_1 \\
+d_5 &= pd_4 + qd_3 + rd_2 \\
+d_6 &= pd_5 + qd_4 + rd_3
+\end{align*}
+</cmath>
+This is a set of three linear equations. In our case, it has a unique solution </math>(p,q,r)=(-2,-1,1)<math>, hence </math>d_7 = -2d_6 - d_5 + d_4 = \boxed{012}$.
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