Solving recurrence relations using characteristics root techinque

Question

During solving recurrence relations using characteristics root techinque why do we assume $F(n) = r^n$?

Answer 1

If we have the equation $$ (S-r_1)(S-r_2)\cdots(S-r_k)\,a_n=0\tag1 $$ where $Sa_n=a_{n+1}$ is the shift operator on sequences, then we have the $k$ solutions to $$ (S-r_j)\,a_n=0\tag2 $$ and since $(2)$ is $a_{n+1}=r_ja_n$, the solution to $(2)$ is pretty clearly $a_n=c_jr_j^n$.

Thus, $$ a_n=\sum_{j=1}^kc_jr_j^n\tag3 $$ is a solution to $(1)$, but since $k$ initial values fix $a_n$, varying the $k$ $c_j$s gives all the solutions.