For a natural number n , let α n = 19 n - 12 n . Then, the value of 31 α 9 - α 10 57 α 8…
For a natural number , let . Then, the value of is ______
Solution
If $r_1$ and $r_2$ are roots of the quadratic $ax^2 + bx + c = 0$
then $r_1^n = 31r_1^{n-2} - 228r_1^{n-1}$ and $r_2^n = 31r_2^{n-2} - 228r_2^{n-1}$
So, $r_1^n - r_2^n = 31(r_1^{n-2} - r_2^{n-2}) - 228(r_1^{n-1} - r_2^{n-1})$
Or $a_n - 31a_{n-1} + 228a_{n-2} = 0$ where $a_n = r_1^n - r_2^n$
Since $19$ and $12$ are the roots of the quadratic $x^2 - 31x + 228 = 0$
we know, $a_n - 31a_{n-1} + 228a_{n-2} = 0$
Now, for $n = 10$, $\alpha_{10} - 31\alpha_{9} + 228\alpha_{8} = 0$
$\Rightarrow \frac{31\alpha_{9} - \alpha_{10}}{57\alpha_{8}} = 4$
Asked in: JEE Main 2022 (25 Jun Shift 1)
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