# Difference between revisions of "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

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