# Thread: Second Order Recurrence relation

1. ## Second Order Recurrence relation

I can usually solve second order ones fine but this one has me stumped!

a(n) = 6a(n-1) - 9a(n-2) a(0) = 1 a(1) = 1

Using t^2 = 6t^(n-1) - 9t^(n-2)
I get t = 3 and only 3

When I try to plug in S(n) = 3^n and T(n) = 3^n and solve for

U(n) = bS(n) + dT^n with b3^n + d3^n and use my initial values of 1 for U(0) and for U(1) I can not find a solution for the values of b and d that meet both of these!!!!!! PLease give me a clue or two or three.....

2. Running it through Maple, I get:

$a_{n}=(1-\frac{2n}{3})\cdot{3^{n}}$

That's something to shoot for.

3. Hello, frostking2!

I think we speak the same language ... We just use different symbols.

$a(n) \:= \:6a(n-1) - 9a(n-2),\quad a(0) = 1,\;\;a(1) = 1$
Here's the way I was taught to handle these . . .

We conjecture that: . $a(n) \:=\:X^n$ . . . that the function is exponential.

The equation becomes: . $X^n \:=\:6X^{n-1} - 9X^{n-2}\quad\Rightarrow\quad X^n - 6X^{n-1} + 9X^{n-2}\;=\;0$

Divide by $X^{n-2}\!:\;\;X^2 - 6X + 9 \:=\:0\quad\Rightarrow\quad (X-3)^2\:=\:0\quad\Rightarrow\quad X \:=\:3,\,3$

With repeated roots, the function is: . $a(n) \;=\;A\!\cdot\!3^n + B\!\cdot\!n\!\cdot\!3^n$

Plug in the first two values of the sequence . . .

. . $a(0) = 1:\;A\!\cdot\!3^0 + B(0)(3^0) \:=\:1 \quad\Rightarrow\quad A \:=\:1$

. . $a(1) = 1:\;A\!\cdot\!3^! + B(1)(3^1) \:=\:1\quad\Rightarrow\quad B \:=\:-\frac{2}{3}$

Hence: . $a(n) \;=\;3^n - \frac{2}{3}n\cdot3^n \;=\;3^n\left(1 - \frac{2}{3}n\right) \;=\;3^n\left(\frac{3-2n}{3}\right)$

Therefore: . $\boxed{a(n) \;=\;3^{n-1}(3 - 2n)}$

4. ## Second order problem solved!

Thank you soooooo much. Stupid me I NEVER considered 0 for one of the values at the last portion of the problem, second roots and values for a and b. Of course it does work and so the regular process yields an answer. Thanks so much for your time and helpful attitude!!!!!