[tex]u_{n+1}=3u_{n}-n-1\Leftrightarrow u_{n+1}-\frac{1}{2}(n+1)-\frac{3}{4}=3\left ( u_{n}-\frac{1}{2}n-\frac{3}{4} \right )[/tex]
Đặt [tex]v_{n}=u_{n}-\frac{1}{2}n-\frac{3}{4}\Rightarrow \left\{\begin{matrix} v_{1}=u_{1}-\frac{1}{2}.1-\frac{3}{4}=-\frac{1}{2} & \\ v_{n+1}=3v_{n} & \end{matrix}\right.[/tex]
[tex]\Rightarrow v_{n}[/tex] là CSN với công bội 3
[tex]\Rightarrow v_{n}=-\frac{1}{2}.3^{n-1}\Rightarrow u_{n}=v_{n}+\frac{1}{2}n+\frac{3}{4}=-\frac{1}{2}3^{n-1}+\frac{n}{2}+\frac{3}{4}[/tex]