3 ms·
No factoring or quadratic formulas necessary. I'm going to write c_n instead of a_n and a,b instead of alpha,beta. Since a and b satisfy that equation, you kno
by nilkn 13y ago
No factoring or quadratic formulas necessary. I'm going to write c_n instead of a_n and a,b instead of alpha,beta.
Since a and b satisfy that equation, you know that
a^2 = 6a+ 2, b^2 = 6b+ 2
and hence
c_10 - 2 * c_8 = a^8 * (6a + 2) - b^8 * (6b + 2) + 2b^8 - 2a^8 = 6c_9.
Dividing by 2c_9 gives 3.
Edit: asafira beat me to it!
- alok-g 13y agoUsing the same convention as yours, I tried to break c_10 in terms of c_9 as follows: (a^10-b^10) = (e a + f b) * (a^9-b^9) Multiplying out the RHS, e = f = 1 are the best choices, but there remained a residual. At a second glance though, this residual was proportional to (a^8-b^8). Problem solved.