Art of Problem Solving

2008 AMC 10B Problems/Problem 11: Difference between revisions

BOGTRO (talk | contribs)
m fmt
Line 1: Line 1:
==Problem==
==Problem==
{{Suppose that <math>(u_n)</math> is a serquence of real numbers satifying <math>u_{n+2}=2u_{n+1}+u_n</math>,
Suppose that <math>(u_n)</math> is a [[sequence]] of real numbers satifying <math>u_{n+2}=2u_{n+1}+u_n</math>,


and that <math>u_3=9</math> and <math>u_6=128</math>. What is <math>u_5</math>?
and that <math>u_3=9</math> and <math>u_6=128</math>. What is <math>u_5</math>?


(A) 40 (B) 53 (C) 68 (D) 88 (E) 104}}
<math>\mathrm{(A)}\ 40\qquad\mathrm{(B)}\ 53\qquad\mathrm{(C)}\ 68\qquad\mathrm{(D)}\ 88\qquad\mathrm{(E)}\ 104</math>


==Solution==
==Solution==
{{We konw that <math>u_6=128</math>, so we plug in <math>n=4</math> to get <math>128=2u_5+u_4</math>. We plug in <math>n=3</math> to get
Plugging in <math>n=4</math>, we get  


<math>u_5=2u_4+9</math>. Substituting gives
<center><math>128=2u_5+u_4.</math></center>


<math>128=5u_4+18 \rightarrow u_4=22</math>
Plugging in <math>n=3</math>, we get


This gives <math>u_5=\frac{128-22}{2}=53</math>.
<center><math>u_5=2u_4+9.</math></center>


Answer B is the correct answer
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>.
 
NOTE: This is my (BOGTRO) solution, not the official one,
 
and should be ignored in view of a better solution.}}


==See also==
==See also==
{{AMC10 box|year=2008|ab=B|num-b=9|num-a=11}}
{{AMC10 box|year=2008|ab=B|num-b=10|num-a=12}}


==See also==
[[Category:Introductory Algebra Problems]]
{{AMC10 box|year=2008|ab=B|num-b=10|num-a=12}}

Revision as of 13:28, 11 August 2008

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 $u_5=\frac{128-22}{2}=53 \Longrightarrow \textbf{(B)}$.

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