Difference between revisions of "Mock AIME 1 Pre 2005 Problems/Problem 13"

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=Solution=
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==Problem==
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A sequence <math> \{R_{n}\}_{n\ge 0} </math> obeys the recurrence <math> 7R_{n}= 64-2R_{n-1}+9R_{n-2} </math>  for any integers <math>n\ge 2</math>.  Additionally, <math>R_0=10</math> and <math>R_1=-2</math>.  Let  <cmath> S =\sum_{i=0}^{\infty}\frac{R_{i}}{2^{i}} </cmath> 
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<math>S</math> can be expressed as <math>\frac{m}{n}</math> for two 2 relatively prime numbers <math>m</math> and <math>n</math>.  Determine the value of <math>m+n</math>
  
First we consider the fact that  
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==Solution==
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First, we consider the fact that  
 
<math> 7R_{n} +2R_{n-1}-9R_{n-2}= 64 </math>
 
<math> 7R_{n} +2R_{n-1}-9R_{n-2}= 64 </math>
  

Latest revision as of 14:22, 26 March 2011

Problem

A sequence $\{R_{n}\}_{n\ge 0}$ obeys the recurrence $7R_{n}= 64-2R_{n-1}+9R_{n-2}$ for any integers $n\ge 2$. Additionally, $R_0=10$ and $R_1=-2$. Let \[S =\sum_{i=0}^{\infty}\frac{R_{i}}{2^{i}}\] $S$ can be expressed as $\frac{m}{n}$ for two 2 relatively prime numbers $m$ and $n$. Determine the value of $m+n$

Solution

First, we consider the fact that $7R_{n} +2R_{n-1}-9R_{n-2}= 64$

Now, consider the fact that

$S= \frac{R_0}{1} +\frac{R_1}{2} +\frac{R_2}{4} + \frac{R_3}{8} ...$

Thus,

$7S= \frac{7R_0}{1} +\frac{7R_1}{2} +\frac{7R_2}{4} + \frac{7R_3}{8}...$

and

$S= \frac{2R_0}{2} +\frac{2R_1}{4} +\frac{2R_2}{8} + \frac{2R_3}{16} ...$

and


$\frac{-9S}{4}= \frac{-9R_0}{4} +\frac{-9R_1}{8} +\frac{-9R_2}{16} + \frac{-9R_3}{32} ...$

Adding these together we get that

$S=\frac{420}{23}$

Since 420 and 23 are relatively prime we find that the answer is $\boxed{443}$