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

Revision as of 16:26, 10 August 2008

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

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

(A) 40 (B) 53 (C) 68 (D) 88 (E) 104

We know that $u_6=128$ , so we plug in $n=4$ to get $128=2u_5+u_4$ . We plug in $n=3$ to get

$u_5=2u_4+9$ . Substituting gives

$128=5u_4+18 \rightarrow u_4=22$

This gives $u_5=\frac{128-22}{2}=53$ .

Answer B is the correct answer

NOTE: This is my (BOGTRO) solution, not the official one,

and should be ignored in view of a better solution.

2008 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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

@@ 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>,
+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
 ==Solution==
-{{solution}}
+We know 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
+<math>u_5=2u_4+9</math>. Substituting gives
+<math>128=5u_4+18 \rightarrow u_4=22</math>
+This gives <math>u_5=\frac{128-22}{2}=53</math>.
+Answer B is the correct answer
+NOTE: This is my (BOGTRO) solution, not the official one,
+and should be ignored in view of a better solution.
 ==See also==
 {{AMC10 box|year=2008|ab=B|num-b=9|num-a=11}}