2008 AMC 10B Problems/Problem 11

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

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

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2008 AMC 10B Problems/Problem 11

Problem

Solution

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