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Difference between revisions of "2008 AMC 10B Problems/Problem 11"

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==Solution==
 
==Solution==
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Plugging in <math>n=3</math>, we get  
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If we plug in <math>n=4</math>, we get  
  
 
<center><math>128=2u_5+u_4.</math></center>  
 
<center><math>128=2u_5+u_4.</math></center>  
  
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Plugging in <math>n=3</math>, we get  
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By plugging in <math>n=3</math>, we get  
  
 
<center><math>u_5=2u_4+9.</math></center>
 
<center><math>u_5=2u_4+9.</math></center>
  
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This is simply a system of two equations with two unknowns. Substituting gives <math>128=5u_4+18 \Longrightarrow u_4=22</math>, and <math>u_5=\frac{128-22}{2}=53 \Longrightarrow \textbf{(B)}</math>.
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This is a system of two equations with two unknowns. Multiplying the second equation by 2 and substituting into the first equation gives <math>128=5u_4+18 \Longrightarrow u_4=22</math>, therefore <math>u_5=\frac{128-22}{2}=53 \longrightarrow \textbf{\boxed{(B)}}</math>.
  
 
==See also==
 
==See also==

Latest revision as of 15:15, 7 June 2024

Problem

Suppose that $(u_n)$ is a sequence of real numbers satifying $u_{n+2}=2u_{n+1}+u_n$,

and that $u_3=9$ and $u_6=128$. What is $u_5$?

$\mathrm{(A)}\ 40\qquad\mathrm{(B)}\ 53\qquad\mathrm{(C)}\ 68\qquad\mathrm{(D)}\ 88\qquad\mathrm{(E)}\ 104$

Solution

If we plug in $n=4$, we get

$128=2u_5+u_4.$

By plugging in $n=3$, we get

$u_5=2u_4+9.$

This is a system of two equations with two unknowns. Multiplying the second equation by 2 and substituting into the first equation gives $128=5u_4+18 \Longrightarrow u_4=22$, therefore $u_5=\frac{128-22}{2}=53 \longrightarrow \textbf{\boxed{(B)}}$.

See also

2008 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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 10 Problems and Solutions

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