Solving general solution of recurrence relation by iteration

Question

$$a_{n-1} = ca_{n-2} $$

Hence

$$a_n = c \cdot c \cdot a_{n-2} $$
$$ = c \cdot c \cdot c \cdot a_{n-3} $$
......

$$ = c^na_0 $$

Why is there a iteration on the constant $c$ ?

Answer 1

Your basic equation says that to get from $a_{n-2}$ to $a_{n-1}$ you multiply by $c$. If one step corresponds to multiplying by $c$, two steps should be multiplying by $c$ twice, which is $c\cdot c=c^2$ and $n$ steps should be multiplying by $c^n$.

An explicit example: take $a_0=1, c=2.$ Then your first equation says (taking $n=1$) $a_1=2a_0=2$, Then (taking $n=2$) $a_2=2a_1=4$ Does this help?