For a natural number n , let α _ n = 19 ^ n - 12 ^ n . Then, the...

If ${r}_{1}$ and ${r}_{2}$ are roots of the quadratic $a{x}^{2} + bx + c = 0$

then ${{r}_{1}}^{n} = 31{{r}_{1}}^{n - 2} - 228{{r}_{1}}^{n - 1}$ and ${{r}_{2}}^{n} = 31{{r}_{2}}^{n - 2} - 228{{r}_{2}}^{n - 1}$

So, ${{r}_{1}}^{n} - {{r}_{2}}^{n} = 31\left({{r}_{1}}^{n - 2} - {{r}_{2}}^{n - 2}\right) - 228\left({{r}_{1}}^{n - 1} - {{r}_{2}}^{n - 1}\right)$

Or ${a}_{n} - 31{a}_{n - 1} + 228{a}_{n - 2} = 0$ where ${a}_{n} = {{r}_{1}}^{n} - {{r}_{2}}^{n}$

Since $19 \text{ and } 12$ are the roots of the quadratic ${x}^{2} - 31x + 228 = 0$

we know, ${a}_{n} - 31{a}_{n - 1} + 228{a}_{n - 2} = 0$

Now, for $n = 10, {\alpha}_{10} - 31{\alpha}_{3} + 228{\alpha}_{8} = 0$

$⇒ \dfrac{31{\alpha}_{9} - {\alpha}_{10}}{57{\alpha}_{8}} = 4$