Quadratic Equations
Newton's Sums
JEE Advanced 2011
Grade 11
Question:
Let $\alpha > \beta$ be the roots of $x^2 - 6x - 2 = 0$. If $a_n = \alpha^n - \beta^n$ for $n \geq 1$, then $\frac{a_{10} - 2a_8}{2a_9}$ equals:
(1) $1$
(2) $2$
(3) $3$
(4) $4$
Step-by-Step Solution
Key Concept: Since $\alpha^2 = 6\alpha + 2$ (root), replace $\alpha^{10} = \alpha^8(\alpha^2) = \alpha^8(6\alpha + 2)$. Do the same for $\beta^8$. The combination $a_{10} - 2a_8$ telescopes neatly.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (3)