Solve the recurrence relation: $a_n = 6a_{n-1} - 8a_{n-2}; a_0 = 1, a_1 = 0 $

Question

anonymous · Answer

$a_2 = -8, a_3 = -48, a_4 = -224\
x^n = 6x^{n-1} - 8x^{n-2} \equiv x^n - 6x^{n-1} + 8x^{n-2} = 0\
x^2 - 6x + 8 = 0; x = 6,4$

So now I should just be able to find constants to fulfill $a_n = A6^n + B4^n$ ?

anonymous · Answer

Not sure how you got that second equaation

anonymous · Answer

is this linear algebra?

xapproachesinfinity · Answer

Me neither!

anonymous · Answer

Discrete math. For linear homogeneous recurrence relations with constant coefficients you're supposed to be able to find a solution of the form $S_n = t^n$ which comes out to that.

anonymous · Answer

Yeah, just solve for $A$ and $B$ and see if it works

anonymous · Answer

ok yeah it works