First-order linear recurrence ($u_{k+1}=a+b{u}_k$)

For $b\ne 1$, subtract the equilibrium $u_{\ast }=a/(1-b)$ to get $u_{k+1}-u_{\ast }=b(u_k-u_{\ast })$. Hence $u_k=u_{\ast }+b^k(u_0-u_{\ast })$. For $b=1$, the solution is $u_k=u_0+ka$. The exceptional factor $0^0$ in a formula with $b=0,k=0$ is avoided by retaining the given initial value.