Running it through Maple, I get:
That's something to shoot for.
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.....
Hello, frostking2!
I think we speak the same language ... We just use different symbols.
Here's the way I was taught to handle these . . .
We conjecture that: . . . . that the function is exponential.
The equation becomes: .
Divide by
With repeated roots, the function is: .
Plug in the first two values of the sequence . . .
. .
. .
Hence: .
Therefore: .
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!!!!!