# Questions about using iteration to express a recurisve sequence

1. Nov 24, 2006

### mr_coffee

Hello everyone, i'm having some issues on a few problems and i'm not sure if i did the first one i'm going to post right, any help or clarifcation would be great.

The directions are: In eavch of 3-15 a sequence is defined recursively. Use iteration to guess an explicit formula for the sequence.

and here is my work:

For the last one, I see that the 2 is increasing at the same rate as the 3^k, but i have no idea about the pattern, i'm not sure what technique i'm suppose to use, i know they said "guess" a formula, but what will help me make an educated guess? Of course there are some easy ones like n! that i can recongize but these arn't so easy for me to see whats going on.

I can see i can factor out a 2 and a 3^2 from p_4 but i don't see how that is going to help me much, getting:
$$2^4 + 3^4 + 2*3^2(1 + 3)$$

thanks!

2. Nov 25, 2006

### durt

You made a bit of an arithmetic mistake: that should be $$2^2 3^2$$, not $$2^1 3^2$$. If you continue on I think you'll notice that $$p_n = 2^n + 3^n + \left( 2^{n-2} 3^2 + 2^{n-3} 3^3 +...+ 2^1 3^{n-1} \right)$$. That stuff in parentheses is a geometric series.

3. Nov 26, 2006

### mr_coffee

Thanks durt!
I'm alittle confused on how you got some of the terms of n
$$p_n = 2^n + 3^n + \left( 2^{n-2} 3^2 + 2^{n-3} 3^3 +...+ 2^1 3^{n-1} \right)$$

But the parts of it I don't understand is how do you know after looking at a few terms, like a_1 to a_4 how you figure out which exponents go to a certian power, such as $$2^{n-2} 3^2 + 2^{n-3} 3^3 +...+ 2^1 3^{n-1}$$ ?

I see that $$2^n + 3^n$$ would be just n because they are increasing at the same time, but from there whats the process to break that down into the other powers of n?

Thanks

Also I could replace that geometric series with the following sum right?
First term is $$2^{n-2} 3^2$$
ratio is: 3/2 I believe because 2 is decreasing and 3 is increasing as you progress
term after the last term is: $$3/2* 2^1 3^{n-1}$$ = $$3^{n-2}$$

so now applying a formula:
[mythical next term - first term]/(ratio-1)
= ( $$3^{n-2}$$ - $$2^{n-2} 3^2$$ )/(3/2 -1)

Last edited: Nov 26, 2006