MHB Recurrence Relations - Determining a solution of the recurrence relation

[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 {a_n} a solution of the recurrence relation
a_n = 8a_n-1 - 16a_n-2 if:"

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

My approach to question c

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

 

[skeeter]

a_n = 8a_n-1 - 16a_n-2

(c)

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

 

[bigpunz04]

(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 {a_n} 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(4^n-1) - 16(4^n-2)
=2³(4^n-1) - 2⁴(4^n-2)
=2³(2^2+n-1) - 2⁴(2^2+n-2)
=0

 

[skeeter]

8(4^n-1) - 16(4^n-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
​

 

[bigpunz04]

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

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

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

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

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.

 

[skeeter]

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

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

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

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

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