Recurrence Relation

**jennifer1004** · January 13th 2009, 02:21 PM

Show that the sequence ${a_{n}}$ is a solution of the recurrence relation $a_{n}=-3a_{n-1}+4a_{n-1}$ if $a_{n}=0$

So far, I plugged in $a_{n}=0$ to get $a_{n}=-3(0)+4(0)=0$ to show the sequence is a solution, but I think I'm missing a step. How can I figure out what $a_{n-1}$ and $a_{n-2}$ are?

**Danny** · January 13th 2009, 02:40 PM

Originally Posted by jennifer1004

Show that the sequence ${a_{n}}$ is a solution of the recurrence relation $a_{n}=-3a_{n-1}+4a_{n-1}$ if $a_{n}=0$

So far, I plugged in $a_{n}=0$ to get $a_{n}=-3(0)+4(0)=0$ to show the sequence is a solution, but I think I'm missing a step. How can I figure out what $a_{n-1}$ and $a_{n-2}$ are?

From what it looks like the two terms on the RHS can be added together to give

$a_n = a_{n-1}$ .

Is your right hand side right or is it

$a_{n}=-3a_{n-1}+4a_{n-2}$ ?

**jennifer1004** · January 13th 2009, 02:42 PM

You're right. I'm sorry about that. it is $4a_{n-2}$ .

The edited question is:

Show that the sequence

is a solution of the recurrence relation $a_{n}=-3a_{n-1}+4a_{n-2}$

**Danny** · January 13th 2009, 03:05 PM

Originally Posted by jennifer1004

You're right. I'm sorry about that. it is $4a_{n-2}$ .

Typically given

$a_{n}=-3a_{n-1}+4a_{n-2}$ for $n \ge 2$

you are given $a_0\; \text{and}\; a_1$ , then the recurrance relationship gives the rest of the $a_n 's$ . Is your problem different?

**jennifer1004** · January 13th 2009, 03:11 PM

In other problems I was given the first two cases, but not this one. The entire question is:

Show that the sequence { $a_n$ } is a solution of the recurrence relation $a_{n}=-3a_{n-1}+4a_{n-2}$ if $a_{n}=0$

**Danny** · January 13th 2009, 03:33 PM

Originally Posted by jennifer1004

In other problems I was given the first two cases, but not this one. The entire question is:

Show that the sequence { $a_n$ } is a solution of the recurrence relation $a_{n}=-3a_{n-1}+4a_{n-2}$ if $a_{n}=0$

Well, you got me. The $a_n = 0$ is bothersome as it specifies nothing of the value of n. Of course the sequence work for $a_n=0$ for all n.

For what it's worth. The recursion equation has the exact solution

$a_n = c_1 + c_2(-4)^n$ for arbitrary constants $c_1,\;\;c_2$

Sorry I couldn't have been of more help.

**jennifer1004** · January 13th 2009, 03:37 PM

Thank you for your help. I was stumped as well because there does not seem to be enough information to work with. I have more of these questions, but hopefully I can post one that will make sense. Thanks again for trying to make sense out of nothing!

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