Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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 $c$ ?

share|improve this question
    
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
    
This page might help you: en.wikipedia.org/wiki/Geometric_progression –  Nick Strehlke May 1 '11 at 2:35
add comment

1 Answer

up vote 2 down vote accepted

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?

share|improve this answer
    
$$a_{n-1} = ca_{n-2} $$ oh ok i got it. –  optimus May 1 '11 at 2:47
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.