I got the hard part of the problem done with help from another forum member,
and i transfered it into a gemeotric sum,
which i will show below but i'm confused on the following
directions that my profressor wrote:

I checked in the section he was talking about and to my luck, no problem
deals with what he is speaking of...but i do have an example of what he means.

In #8 on the image below, i have boxed the check in which he requires. #11. is where i'm stuck,
i'm not sure what i'm suppose to plug in, following his pattern for #8, it looks like he took
the right hand side of the given equation, and set it equal to the formula i found in replace
of f_{k-1} because there is a +2^k after the formula i found in #8 which was: f_n = 2^{n+1} - 3.

Sorry if this is confusing but the directions to these problems are the following:

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

Not quite what you want. You want to have each exponent on the right hand side to be to the power k-1 (since you are looking at [itex]p_{k-1}[/itex]). Also you want 2 times that, so you need to factor the 2 out. So you should have:
[tex]2p_{k-1} + 3^k = 2 \left(2^{k-1} + 3^{k-1} + \frac{4\cdot 3^{k-1} - 9\cdot 2^{k-1}}{2} \right) + 3^k[/tex]

Maybe I showed a little too much work there, but you should notice that I could have "simplified" things, like 2x9 = 18, but we do not always want to do that (remember that we know what we want our answer to be, so try to get the answer to look like that).

Actually, your answer of 3*3^k - (7*2^k)/2 is p_k.

As I said above, you have a goal that you want to reach so try to do the algebra with that goal in mind.

[tex]3\cdot3^k - \frac{7\cdot 2^k}{2} = 3^k + 2\cdot 3^k - \frac{7\cdot 2^k}{2}[/tex]
[tex]= 3^k + \frac{4\cdot 3^k - 7\cdot 2^k}{2}[/tex] multiply by 2/2 and then put as a bigger fraction
[tex]= 3^k + \frac{4\cdot 3^k - 9\cdot 2^k + 2\cdot 2^k}{2}[/tex] add "0" subtract 2*2^k then add 2*2^k
[tex]= 3^k + \frac{4\cdot 3^k - 9\cdot 2^k}{2} + \frac{2\cdot 2^k}{2}[/tex] split up fraction
[tex]= 2^k + 3^k + \frac{4\cdot 3^k - 9\cdot 2^k}{2}[/tex]
[tex]= p_k[/tex]

All he is asking is that when you guess a solution you do two things.

1. Check that the guess is correct for the base case
2. Check that the guess actually satisfies the recurrence relation

I.e. things you should have done anyway but which the teacher is obviously fed up of students ignoring and so is making you remember to do it by this unsubtle but understandable tactic.

I understand his reasonings i just was confused on how he was subbing in the explicit formula into the orginal formula, the section he talks of didn't have an example of this kind but used induction rather than this technique.

You are actually doing induction, just in a slightly different way.

First you are showing that the first case(s) are true (think of base case(s)), and then you are showing that if the formula is true for k-1, then it is also true for k (the inductive step).