Let α and β Be the Roots of the Equation x² − 6x − 2 = 0. If aₙ = αⁿ − βⁿ for n ≥ 1, Then the Value of (a₁₀ − 2a₈)/(2a₉) Is Equal To

JEE Main 2015. Previous Year Question.

Problem

Let $\alpha$ and $\beta$ be the roots of the equation $x^2-6x-2=0$. If $a_n=\alpha^n-\beta^n$, for $n\geq1$, then the value of $\dfrac{a_{10}-2a_8}{2a_9}$ is equal to:

Key insight. Since $\alpha$ and $\beta$ satisfy the same quadratic, multiplying that equation by $\alpha^{n-1}$ for $\alpha$ and by $\beta^{n-1}$ for $\beta$ and subtracting produces a recurrence relating $a_{n+1}$, $a_n$, and $a_{n-1}$ directly from the quadratic’s coefficients — Newton’s formula for power sums. This sidesteps ever needing to know $\alpha$ or $\beta$ individually.

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Approach

Since $\alpha,\beta$ are roots of $ax^2+bx+c=0$, they satisfy $a\alpha^2+b\alpha+c=0$ and $a\beta^2+b\beta+c=0$. Multiplying by $\alpha^{n-1}$ and $\beta^{n-1}$ respectively and subtracting gives the recurrence $a\cdot a_{n+1}+b\cdot a_n+c\cdot a_{n-1}=0$. Apply this with $n=9$ using the given quadratic’s coefficients, then rearrange to match the required expression.

Solution

Step 1 — Identify the coefficients

For $x^2-6x-2=0$: $a=1$, $b=-6$, $c=-2$.

Step 2 — Derive the recurrence

Since $\alpha^2-6\alpha-2=0$ and $\beta^2-6\beta-2=0$, multiplying by $\alpha^{n-1}$ and $\beta^{n-1}$ respectively and subtracting:

$$(\alpha^{n+1}-\beta^{n+1}) – 6(\alpha^n-\beta^n) – 2(\alpha^{n-1}-\beta^{n-1}) = 0$$

$$a_{n+1} – 6a_n – 2a_{n-1} = 0$$

Step 3 — Apply the recurrence with n = 9

$$a_{10} – 6a_9 – 2a_8 = 0$$

Step 4 — Rearrange to isolate a10 − 2a8

$$a_{10} – 2a_8 = 6a_9$$

Step 5 — Divide both sides by 2a9

$$\frac{a_{10}-2a_8}{2a_9} = \frac{6a_9}{2a_9} = 3$$

Answer

Common mistakes

• Trying to compute $\alpha$ and $\beta$ explicitly and raise them to the 8th, 9th, and 10th powers. This is technically possible using the quadratic formula but leads to extremely messy surd arithmetic — the recurrence relation avoids ever finding $\alpha$ or $\beta$ individually.
• Getting a sign wrong when deriving the recurrence. Since the quadratic has $c=-2$ (negative), the recurrence term is $-2a_{n-1}$, not $+2a_{n-1}$ — this sign directly determines whether the final rearrangement gives $+6a_9$ or $-6a_9$.

Practise next

• Let $\gamma,\delta$ be the roots of $x^2-4x-1=0$, and $b_n=\gamma^n-\delta^n$. Find $\dfrac{b_8-b_6}{b_7}$, using the same Newton’s-recurrence approach.