# Math Help - Sequences

1. ## Sequences

Given the sequence:

u0=3 u1=22

un+2= (12un+1)-(20un)

To find the closed for we use the equation

r^2-12r+20=0

So, why do we use that equation, how does it work? Why are those signs switched, so positive becomes a negative and vice versa?

I know the answer and how to get it, so

un=2^n+2x10^10

I just need some help on how all this works.

Thanks.

2. Originally Posted by mark090480
Given the sequence:

u0=3 u1=22

un+2= (12un+1)-(20un)

To find the closed for we use the equation

r^2-12r+20=0

So, why do we use that equation, how does it work? Why are those signs switched, so positive becomes a negative and vice versa?

I know the answer and how to get it, so

un=2^n+2x10^10

I just need some help on how all this works.

Thanks.
Hi Mark,

for $U_{n+2}=12U_{n+1}-20U_n$

then since they are equal, subtract them and the answer is zero.

$U_{n+2}-12U_{n+1}+20U_n=0$

then write the characteristic equation and get the roots $\alpha$ and $\beta$

using $U_{n+2}\rightarrow\ r^2,\ U_{n+1}\rightarrow\ r^1,\ U_n\rightarrow\ r^0=1$

$r^2-12r+20=(r-10)(r-2)=0$

$\alpha=10,\ \beta=2$

Then you use your values of $U_1$ and $U_0$ to finish up.

3. Hello, mark090480!

Here's a primitive explanation . . .

Given the sequence: . $U_{n+2} \:=\:12\,U_{n+1} - 20\,U_n,\;\;\;U_0 = 3,\;U_1 = 22$

To find the closed for we use the equation . $r^2-12r + 20 \:=\:0$

So why do we use that equation, how does it work?

We have: . $U_{n+2} \:=\:12\,U_{n+1} - 20\,U_n$

Assume that $U_n$ is an exponential function: . $U_n \:=\:r^n$

Then we have: . $r^{n+2} \:=\:12r^{n+1} - 20r^n \quad\Rightarrow\quad r^{n+2} - 12r^{n+1} + 20r^n \:=\:0$

Divide by $r^n\!:\;\;\;r^2 - 12r + 20 \:=\:0$

Got it?

4. OK thanks, I get it I think.

so, if we assume the Fibonacci sequence

5= 3+2

5-3-2=0

hence that is why. I'm not sure still as to how solving that equation gives us the closed form, could you explain that?