First-order linear recurrence
= First-order linear recurrence
{title2=$u_{k+1}=a+bu_k$}
For $b\ne1$, subtract the equilibrium $u_*=a/(1-b)$ to get $u_{k+1}-u_*=b(u_k-u_*)$. Hence $u_k=u_*+b^k(u_0-u_*)$. 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.