Problem 159 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(\alpha\) and \(\beta\) be the roots of equation \(x^{2}-6 x-2=0\). If \(a_{n}=\alpha^{n}-\beta^{n}\), for \(n \geq 1\),then the value of \(\frac{a_{10}-2 a_{8}}{2 a_{9}}\) is equal to (A) \(-6\) (B) 3 (C) \(-3\) (D) 6

Step-by-Step Solution

Answer

The value is 3.

Step 1: Identify Relationship Using Vieta's Formulas

From the equation \(x^2 - 6x - 2 = 0\), the sum of the roots \(\alpha + \beta = 6\) and the product \(\alpha\beta = -2\) are derived.

Step 2: Establish Recursive Relation

The sequence \(a_n = \alpha^n - \beta^n\) satisfies the recurrence \(a_n = (alpha + \beta) a_{n-1} - \alpha\beta a_{n-2}\). Substituting the values from Step 1, we get \(a_n = 6a_{n-1} + 2a_{n-2}\).

Step 3: Compute Initial Terms

Calculate \(a_1\) and \(a_2\). We have \(a_1 = \alpha - \beta\) and \(a_2 = \alpha^2 - \beta^2 = (\alpha + \beta)(\alpha - \beta) = 6(\alpha - \beta)\). Thus, \(a_1 = \alpha - \beta\), \(a_2 = 6a_1\).

Step 4: Calculate Higher Terms

Use the recursive relation \(a_n = 6a_{n-1} + 2a_{n-2}\) to compute \(a_8, a_9,\) and \(a_{10}\), though exact terms are not needed as we will use the recurrence relation to substitute into our expression directly.

Step 5: Simplify Expression

Use the recursive relation to simplify the expression \(\frac{a_{10} - 2a_8}{2a_9} = \frac{(6a_9 + 2a_8) - 2a_8}{2a_9} = \frac{6a_9}{2a_9} = 3\). Therefore, the expression simplifies to 3.

Key Concepts

Vieta's Formulas

Vieta's Formulas give us a profound connection between the coefficients of a polynomial equation and its roots. For a quadratic equation of the form \(ax^2 + bx + c = 0\), Vieta's Formulas state:

• The sum of the roots \(\alpha + \beta = -\frac{b}{a}\)
• The product of the roots \(\alpha \beta = \frac{c}{a}\)

Using the equation from our exercise, \(x^2 - 6x - 2 = 0\), which can be rewritten as \(1\cdot x^2 - 6\cdot x - 2 = 0\), we can easily apply these formulas:

• Sum: \(\alpha + \beta = 6\)
• Product: \(\alpha \beta = -2\)

These outcomes help us establish relationships between the roots that are critical in solving problems dealing with powers of roots or sequences derived from them.
Understanding Vieta's Formulas can simplify the process of dealing with roots without solving the equations explicitly, saving time and effort.

Recursive Relations

Recursive relations provide a systematic way to construct terms of a sequence step by step. In this exercise, we are given the sequence \(a_n = \alpha^n - \beta^n\), and our task is to identify the recursive relationship that the sequence satisfies.
Using the values from Vieta's Formulas, the relationship can be expressed as \(a_n = (\alpha+\beta)a_{n-1} - \alpha\beta a_{n-2}\). In this specific case, it simplifies to:

• \(a_n = 6a_{n-1} + 2a_{n-2}\)

To start the sequence, we need the initial terms, which we compute as:

• \(a_1 = \alpha - \beta\)
• \(a_2 = 6a_1\)

This relation allows us to generate any term of the sequence using only the two preceding terms. By recognizing and using recursive relations, calculations become more manageable, particularly for larger \(n\) or when expressions become complex.

Sequence Calculation

Once we have a recursive relation for a sequence, calculating specific terms becomes a step-by-step process. For the sequence \(a_n = \alpha^n - \beta^n\), which adheres to \(a_n = 6a_{n-1} + 2a_{n-2}\), we can calculate terms progressively.
To solve the given problem, though, we can use the recurrence relation directly in the expression \(\frac{a_{10} - 2a_8}{2a_9}\). By applying the recursive relation:

• \(a_{10} = 6a_9 + 2a_8\)
• \(a_9 = 6a_8 + 2a_7\)

At this stage, focus on how the expression simplifies:

• \(\frac{a_{10} - 2a_8}{2a_9} = \frac{(6a_9 + 2a_8) - 2a_8}{2a_9} = \frac{6a_9}{2a_9} = 3\)

The arithmetic simplifies due to the relation, allowing us to find the expression's value without explicitly calculating \(a_8, a_9, \) or \(a_{10}\). Using recursive relations creatively enables efficient problem-solving and deeper understanding of sequences, reinforcing mathematical insight into the underlying patterns.