SOLUTION: If the series {ak} satisfies that a1 = 1, and a2 = 2, and ak - 4a(k-1) + 3a(k-2) = 0 (k >= 0), then ak = (1 + p)/q. Find p and q for all values of k >= 1.
Algebra ->
Sequences-and-series
-> SOLUTION: If the series {ak} satisfies that a1 = 1, and a2 = 2, and ak - 4a(k-1) + 3a(k-2) = 0 (k >= 0), then ak = (1 + p)/q. Find p and q for all values of k >= 1.
Log On
Question 1163040: If the series {ak} satisfies that a1 = 1, and a2 = 2, and ak - 4a(k-1) + 3a(k-2) = 0 (k >= 0), then ak = (1 + p)/q. Find p and q for all values of k >= 1. Found 2 solutions by Edwin McCravy, ikleyn:Answer by Edwin McCravy(20056) (Show Source):
Something is wrong with this problem, as I show below.
If the series {ak} satisfies that a1 = 1, and a2 = 2, and ak - 4a(k-1) + 3a(k-2) = 0 (k >= 0), then ak = (1 + p)/q. Find p and q for all values of k >= 1.
We substitute k=2
-----
Next we substitute k=3
So we have these two equations in p and q:
This can't be because the solution to that system is p=-1, q=0,
But q is in the denominator of
and so q cannot be 0. So the problem is botched.
You can correct it in the space below if you like
and I'll get back to you by email.
Edwin
