1. simplify recurrence relation

I'm trying to solve this for b:

b*2^n = 3*b*2^(n-1) + 2^n

b = -2^(n + 1)

but i don't see how to get there.

can someone enlighten me?

thanks

dan mcleran

2. Originally Posted by danmcleran
I'm trying to solve this for b:

b*2^n = 3*b*2^(n-1) + 2^n

b = -2^(n + 1)

but i don't see how to get there.

can someone enlighten me?

thanks

dan mcleran
$\displaystyle b2^n = 3b2^{n-1} + 2^n$ is not a recurrence relation. It is an equation in b and n.
Solving it gives $\displaystyle b = -2$.

Can you post the complete question? Maybe then, we can see why $\displaystyle b = -2^{n + 1}$ is the correct answer.

3. Full problem

an = 3an-1 + 2^nwith a1 = 5

I was able to solve the homogeneous part to get:

f(n) = a3^n

I then substituted the particular solution into the above recurrence:

b•2^n = 3 b•2^(n-1) + 2^n

But I think I am too rusty at the algebra to solve for b. I was really looking for some algebra help I guess.

Thanks

Dan

4. Hello, Dan!

$\displaystyle a_n \:=\: 3a_{n-1} + 2^n,\quad a_1 = 5$

I was able to solve the homogeneous part to get: .$\displaystyle f(n) \:=\: a\!\cdot\!3^n$

I then substituted the particular solution into the above recurrence:
. . $\displaystyle b\cdot 2^n \:=\: 3b \cdot 2^{n-1} + 2^n$

But I think I am too rusty at the algebra to solve for $\displaystyle b.$
As we say in the Marines, "Ya done good!"

We have: .$\displaystyle b\!\cdot\!2^n \:=\: 3b\!\cdot\!2^{n-1} + 2^n$

Divide by $\displaystyle 2^{n-1}\!:\;\;2b \:=\:3b + 2 \quad\rightarrow\quad b \:=\:-2$

Hence: .$\displaystyle f(n) \;=\;a\!\cdot\!3^n - 2\!\cdot\!2^n$

Since $\displaystyle a_1 = 5$, we have: .$\displaystyle a\!\cdot\!3 - 2^2 \:=\:5 \quad\Rightarrow\quad a \,=\,3$

. . Hence: .$\displaystyle f(n) \;=\;3\!\cdot\!3^n - 2\!\cdot\!2^n$

Therefore: .$\displaystyle f(n) \;=\;3^{n+1} - 2^{n+1}$

5. I see now. Thanks a lot.