Suppose that $(u_n)$ is a [sequence](https://artofproblemsolving.com/wiki/index.php/Sequence) of real numbers satifying $u_{n+2}=2u_{n+1}+u_n$,
and that $u_3=9$ and $u_6=128$. What is $u_5$?
$\mathrm{(A)}\ 40\qquad\mathrm{(B)}\ 53\qquad\mathrm{(C)}\ 68\qquad\mathrm{(D)}\ 88\qquad\mathrm{(E)}\ 104$
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 $u_5=\frac{128-22}{2}=53 \longrightarrow \textbf{\boxed{(B)}}$.
~ Mathkiddie