and
For "linear recurrences with constant coefficients", things such as where b and c are numbers, there is a very general method: first drop the number that is not multiplying and "a": and "try" . Then so the equation becomes and, dividing both sides by , r= b. That is satisfies . In fact it is easy to see that any constant times that, , is a solution. To find a solution to the entire equation, "try" a constant: if the equation becomes so and (if b is not 1), . The solution to the entire equation is the sum of those: for any constant p. (If b= 1, the equation is and we have an arithmetic sequence.)
This same idea can be used for higher order recurrences: If we have , setting gives [tex]r^{n+2}= 3r^{n+1}- 2r^n[tex] and dividing through by and moving everything to the left, . That is, both and satisfy the recurrence and the general solution is for any constants, A and B.