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$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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    This page might help you: http://en.wikipedia.org/wiki/Geometric_progression2011-05-01

1 Answers 1

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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?

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    $a_{n-1} = ca_{n-2} $ oh ok i got it.2011-05-01