# I got explicit formula to the sequence correct, but confused on checking

• mr_coffee
In summary: Also you want 2 times that, so you need to factor the 2 out. So you should have: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
mr_coffee
Hello everyone.

I got the hard part of the problem done with help from another forum member,
and i transferred it into a gemeotric sum,
which i will show below but I'm confused on the following
directions that my profressor wrote:
Check your answers to 7, 8 and 11 by verifying that your formula satisfies the initial conditions,
and by checking that it satisfies the recurrence relation by substituting it into the recurrence
relation as in Exercise 8.1.9. Include the check as part of your write-up.
Your answer is incomplete if you turn it in without the check, or if you
turn it in even though it does not pass the check. An incomplete answer to 7, 8 or 11
does not check out, then fix it.

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.

and here is my work:

http://suprfile.com/src/1/4ni4e7l/lastscan.jpg Thanks!

Last edited by a moderator:
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 $p_{k-1}$). Also you want 2 times that, so you need to factor the 2 out. So you should have:
$$2p_{k-1} + 3^k = 2 \left(2^{k-1} + 3^{k-1} + \frac{4\cdot 3^{k-1} - 9\cdot 2^{k-1}}{2} \right) + 3^k$$

If you then do some algebra, it all works out.

Last edited:
Thanks matt,

So it looks like, I have to get that right hand side to simplify down to the formula i found which is: P_k = 2^k + 3^k + [4*3^k-9*2^k]/2

By looking at that, do you think that is possible or am I misunderstanding somthing?

Because in #8, that's all he did, was he made the right hand side equal the formula i found.

That is exactly what you want to do, and yes it does work out (keep your goal in mind when doing the algebra).

Thanks i'll work on it and let you know how it goes

i ended up getting:
3*3^k - (7*2^k)/2

I think I'm approaching this wrong,
its not even close to:
2^k + 3^k + [4*3^k-9*2^k]/2

Hmm, not sure how you got that. Let us start from the beginning:

We have the following:

$$2p_{k-1} + 3^k = 2 \left(2^{k-1} + 3^{k-1} + \frac{4\cdot 3^{k-1} - 9\cdot 2^{k-1}}{2} \right) + 3^k$$
$$=2\cdot 2^{k-1} + 2\cdot 3^{k-1} + 4\cdot 3^{k-1} - 9\cdot 2^{k-1}} + 3^k$$ (Just distributing the 2)
$$=2^k + 2\cdot 3^{k-1} + 4\cdot 3^{k-1} - 9\cdot 2^{k-1}} + 3\cdot 3^{k-1}$$ (exponent properties)
$$=2^k + (2+3)\cdot 3^{k-1} + 4\cdot 3^{k-1} - 9\cdot 2^{k-1}}$$ (factoring)
$$=2^k + 3^{k} + 2\cdot 3^{k-1} + 4\cdot 3^{k-1} - 9\cdot 2^{k-1}}$$(distribute the 2 and 3, exponent properties)
$$=2^k + 3^{k} + 6\cdot 3^{k-1} - 9\cdot 2^{k-1}}$$ (group the 2 and 4 from the 3^(k-1))
$$=2^k + 3^{k} + 2\cdot3\cdot 3^{k-1} - 9\cdot 2^{k-1}}$$ (6 = 2x3)
$$=2^k + 3^{k} + 2\cdot 3^{k} - 9\cdot 2^{k-1}}$$ (exponent properties)
$$=2^k + 3^{k} + \frac{2}{2}\left(2\cdot 3^{k} - 9\cdot 2^{k-1}}\right)$$ (multiply by 1)
$$=2^k + 3^{k} + \frac{4\cdot 3^{k} - 9\cdot 2\cdot 2^{k-1}}{2}$$ (distribute 2 on top)
$$=2^k + 3^{k} + \frac{4\cdot 3^{k} - 9\cdot 2^{k}}{2}$$ (exponent properties)
$$=p_k$$

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.

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

Last edited:
Awesome explanation, thanks so much, nice review too.

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).

Oh! that does make sense, never thought of it that way, awesome

## What does it mean to have an explicit formula for a sequence?

An explicit formula for a sequence is a mathematical equation that can be used to calculate any term in the sequence based on its position or index. It is also known as a closed-form formula because it provides a direct way to find any term without having to go through each preceding term.

## Why is it important to have an explicit formula for a sequence?

Having an explicit formula for a sequence can make it easier to find specific terms without having to manually calculate each term before it. It can also help in understanding the pattern or rule behind the sequence and making predictions about future terms.

## How do I know if my explicit formula for a sequence is correct?

To check if your explicit formula for a sequence is correct, you can plug in different values for the index and compare the resulting terms to the actual terms in the sequence. If they match, then your formula is most likely correct. Additionally, you can also check for consistency by making sure that the formula follows the same pattern for different sections of the sequence.

## What should I do if I am confused about checking my explicit formula for a sequence?

If you are confused about checking your explicit formula for a sequence, it may be helpful to consult with a teacher or tutor who can guide you through the process. You can also try using a calculator or computer program to verify your formula and see if it produces the correct results.

## Can an explicit formula for a sequence be wrong?

Yes, an explicit formula for a sequence can be wrong if it does not produce the correct results for all terms in the sequence. This could be due to a mistake in the formula itself or an error in the calculations. It is important to carefully check and verify your formula to ensure its accuracy.

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