Solve the following recurrence relation :

(i) $a_{r}-7 a_{r-1}+10 a_{r-2}=0$ given that $a_{0}=0$, $a_{1}=3$

(ii) $a_{r}-4 a_{r-1}+4 a_{r}-2=0$ given that $a_{0}=1$ and $a_{1}=6$

Solution:

(i) The characteristics equation is,

$ \begin{aligned} \lambda^{2}-7 \lambda+10 &=0 \\\\ (\lambda-5)(\lambda-2) &=0 \\\\ \lambda =5, \lambda=2 \\\\ \end{aligned} $

Therefore the solution of the given recurrence relation is

$ { A_{r}} =\mathrm{A}\left({5^{r}}\right)+\mathrm{B}\left(2^{\mathrm{r}}\right)....(1) \\ $

To where A and B are constants.

To find A and B putting, $r=0$ …