# Solving general solution of recurrence relation by iteration

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

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 $c$ ?

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Please work on formatting your questions with $\LaTeX$ as it makes them more readable. If you look at any formatting on the site and right click, Show Source will be what is inside dollar signs to make it. One set of dollar signs for inline, two for display. –  Ross Millikan May 1 '11 at 2:18
It is not clear to me what your question is. You have shown the solution for $A_n$. This is almost the definition of exponentiation. –  Ross Millikan May 1 '11 at 2:20
hi ross, I dont quite get why for every iteration constant C is added. example in $$a_{n-2}$$ there are 2 c. in $$a_{n-3}$$ there are 3 c –  optimus May 1 '11 at 2:23
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?
$$a_{n-1} = ca_{n-2}$$ oh ok i got it. –  optimus May 1 '11 at 2:47