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

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<center><math>u_5=2u_4+9.</math></center>
 
<center><math>u_5=2u_4+9.</math></center>
  
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 \shortleftarrow \textbf{(B)}</math>.
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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 \longleftarrow \textbf{(B)}</math>.
  
 
==See also==
 
==See also==

Revision as of 21:45, 13 November 2017

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

Plugging in $n=4$, we get

$128=2u_5+u_4.$

Plugging in $n=3$, we get

$u_5=2u_4+9.$

This is simply a system of two equations with two unknowns. Substituting gives $128=5u_4+18 \Longrightarrow u_4=22$, and $u_5=\frac{128-22}{2}=53 \longleftarrow \textbf{(B)}$.

See also

2008 AMC 10B (ProblemsAnswer KeyResources)
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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