Question

Solve a0 = 3; a1 = 6 and an = an-1 + 6an-2; for n greater than or equal to 2.

Answer 1

I will change your an to An to make it easier to read

An = A(n-1) + 6A(n - 2)

An - A(n-1) - 6A(n - 2) = 0

auxiliary equation is

m² - m - 6 = 0

m = 3 or -2

General solution is An = C x 3^n + D x (-2)^n

when n = 0 value is 3 = C + D

when n = 1value is 6 = 3C - 2D


This is what I have so far... Need a solution though?

Answer 2

let me solve this on paper, hold on

Answer 3

Okay great! Thank you so much!!

Answer 4

how did you get your general solution?

Answer 5

Well it may be wrong...I was just brainstorming.

Answer 6

What I did though to get it was I used the form:
\[a _{n} = \alpha6^{n} + \beta (-2)\]

Answer 7

I'm sorry...actually that's wrong. I was taking it from an example problem in the book. The correct formula I used was...

Answer 8

....\[a _{n} = \alpha3^{n} + \beta(-2)^{n}\]

Answer 9

Does that make sense??

Answer 10

It's just the general solution to the recurrence relation that I was formulating...

Answer 11

Then...for n = 0 we needed...
\[3 = a _{0} = \alpha3^{0} + \beta (-2)^{0} = \alpha + \beta \]

Answer 12

And...for n = 1 we needed...
\[6 = a _{1} = \alpha 3^{1} + \beta(-2)^{1} = 3\alpha -2\beta \]

Answer 13

I was just plugging my numbers in...

Answer 14

Oh...so actually that C + D would actually be C - D. I forgot that (-2)^0 would be -1 and not positive 1.

Answer 15

yes, the formula you used makes sense, \[a _{n}=a _{n-1}+6a _{n-2} \]
or, \[\alpha ^{2}-\alpha-6=0 \]
which you had, yielding m=-2, m=3 so the general solution should be \[a _{n}=C3^{n}+D(-2)^{n}\]
if we solve for C and D using the initial conditions above then C=(12/5) and D=(3/5),
plugging in, \[(*) a _{n}=(12/5)3^{n}+(3/5)(-2)^{n}\]
to check: a(2)should = 24=a1+6a0 = 6+6*3=24,
\[a _{2}=(12/5)3^{2}+(3/5)(-2)^{2}=(12/5)9+(3/5)4=24\]
(ah, i made a computational error the first time through, sorry it took a while)
then the general solution is (*) above

Answer 16

I actually got now that
C - D = 3
3C - 2D = 6
Which gives that C = 0 and D = -3

Answer 17

How did you get 12/5 and 3/5?

Answer 18

i used C+D = 3, maybe that was the problem?