# Recurrence Relations - Determining a solution of the recurrence relation

• MHB
• bigpunz04
In summary, the conversation is about understanding how to plug in proposed solutions with n into a recurrence relation. The problem involves finding out if the sequence {an} is a solution of the relation an = 8an-1 - 16an-2. The conversation discusses different solutions (a-e) and how to properly plug them in using the properties of exponents. It is recommended to review these properties in order to better understand the problem.
bigpunz04
Hello -

I am having a tough time understanding the problems in the attached picture (Problem 13). My issue is understanding how I plug in the proposed solutions, specifically those that include n. I am able to solve A and B but unable to solve the rest.

For instance, how do I plug in C or D into the original question? What do I do with the original sub values (n-1) and (n-2) ?

Here is the problem:

"Is the sequence {an} a solution of the recurrence relation
an = 8an-1 - 16an-2 if:"

a) an = 0?
b) an = 1?
c) an = 2n?
d) an = 4n?
e) an = n4n?

My approach to question c

8(2n-1) -16(2n-2)
which I believe gives me...
= 16n-1 - 32n-2

But that answer is obviously wrong. So I'm thinking that I am not plugging the solutions that include n properly. Ugghh so frustrated.

Last edited:
an = 8an-1 - 16an-2

(c)
8(2n-1) -16(2n-2)

note that $2^a(2^b) = 2^{a+b}$ ...

$2^3(2^{n-1}) - 2^4(2^{n-2})$

$2^{n+2} - 2^{n+2} = 0$

use the same property of exponents for (d)

skeeter said:
(c)

note that $2^a(2^b) = 2^{a+b}$ ...

$2^3(2^{n-1}) - 2^4(2^{n-2})$

$2^{n+2} - 2^{n+2} = 0$

use the same property of exponents for (d)

Thank you! That definitely gives me an idea and I will now brush up on my laws of exponents. Haven't used it in a while.

Looking at the answers in the back of the book, #c is not a solution for {an} since it is not equal to 2n. It also mentions that #d is a solution since 4n satisfies the equation. However, when I work the equation the same manner as you did, I end of with 0. It should end up with 4n in order to satisfy the equation.

Here is how I worked the problem. You'll notice that I am stuck again.

=8(4n-1) - 16(4n-2)
=23(4n-1) - 24(4n-2)
=23(22+n-1) - 24(22+n-2)
=0

Last edited:
8(4n-1) - 16(4n-2)

$2^3[(2^2)^{n-1}] - 2^4[(2^2)^{n-2}]$

$2^3[2^{2n-2}] - 2^4[2^{2n-4}]$

$2^{2n+1} - 2^{2n}$

$2^{2n}(2 - 1) = 2^{2n}$

recommend you review properties of exponents

skeeter said:
$2^3[(2^2)^{n-1}] - 2^4[(2^2)^{n-2}]$

$2^3[2^{2n-2}] - 2^4[2^{2n-4}]$

$2^{2n+1} - 2^{2n}$

$2^{2n}(2 - 1) = 2^{2n}$

recommend you review properties of exponents

I think you are right about reviewing exponential properties. Thank you so much.

Last question...

What happened to the +1 where you wrote 22n+1−22n ?

Thanks again, my friend. I now know exactly where I'm lacking and what I need to review.

$2^{2n+1} - 2^{2n}$

$2^{2n} \cdot 2^1 - 2^{2n}$

the two terms above have the common factor $2^{2n}$ ...

$\color{red}{2^{2n}}$ $\cdot 2 -$ $\color{red}{2^{2n}}$

... factor it out from the two terms

$\color{red}{2^{2n}}$ $(2 - 1)$

$2^{2n} (1) = 2^{2n} = 4^n$

## 1. What is a recurrence relation?

A recurrence relation is a mathematical equation that defines a sequence based on the previous terms in the sequence. It is used to model situations where the next term in a sequence depends on the previous terms.

## 2. How do you determine a solution of a recurrence relation?

To determine a solution of a recurrence relation, you can use various methods such as substitution, iteration, or the characteristic equation method. These methods involve finding a pattern in the sequence and using it to find a general formula for the nth term.

## 3. Can all recurrence relations be solved?

No, not all recurrence relations can be solved. Some may have closed-form solutions, while others may only have a recursive definition. In some cases, it may not be possible to find a general formula for the nth term of a sequence.

## 4. How are recurrence relations used in science?

Recurrence relations are used in many scientific fields, including physics, biology, and computer science. They are used to model and analyze complex systems and processes, such as population growth, chemical reactions, and algorithms.

## 5. Are there any real-world applications of recurrence relations?

Yes, recurrence relations have many real-world applications. They are used to predict natural phenomena, optimize processes, and study biological systems. For example, they can be used to predict the population growth of a species or to analyze the efficiency of an algorithm.

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