For a natural number n, let _n=19^n-12^n.

Question

For a natural number $n$, let $\alpha_n=19^n-12^n$. Then the value of $\dfrac{31\alpha_9-\alpha_{10}}{57\alpha_8}$ is \_\_\_

Answer: 4

Accepted Answer

Since $19$ and $12$ are roots of $x^2-31x+228=0$ (as $19+12=31,\ 19\times12=228$), $\alpha_n$ satisfies the recurrence
$$\alpha_{n+1}=31\alpha_n-228\alpha_{n-1}.$$

So $31\alpha_9-\alpha_{10}=31\alpha_9-(31\alpha_9-228\alpha_8)=228\alpha_8$.

$$\frac{31\alpha_9-\alpha_{10}}{57\alpha_8}=\frac{228\alpha_8}{57\alpha_8}=4.$$

Solutions · 1