Let and be the roots of x^2-6x-2=0.

Question

Let $\alpha$ and $\beta$ be the roots of $x^2-6x-2=0$. If $a_n=\alpha^n-\beta^n$ for $n\ge1$, then the value of $\dfrac{a_{10}-2a_8}{3a_9}$ is:

Accepted Answer

Since $\alpha,\beta$ satisfy $x^2-6x-2=0$: $\alpha^2=6\alpha+2$, $\beta^2=6\beta+2$.

$$\frac{a_{10}-2a_8}{3a_9}=\frac{\alpha^8(\alpha^2-2)-\beta^8(\beta^2-2)}{3(\alpha^9-\beta^9)}=\frac{\alpha^8(6\alpha)-\beta^8(6\beta)}{3(\alpha^9-\beta^9)}=\frac{6(\alpha^9-\beta^9)}{3(\alpha^9-\beta^9)}=2.$$

Solutions · 1