imc

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 $128=5u_4+18⟹u_4=22$, therefore $u_5=\frac{128-22}{2}=53⟶\text{53}$.