Problem 159 - A Complete Resource Book in Mathematics for JEE Main 2017

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 \([2015]\) (A) \(-6\) (B) 3 (C) \(-3\) (D) 6

Step-by-Step Solution

Answer

The value is (C) -3.

Step 1: Identify the Roots

First, identify the roots \( \alpha \) and \( \beta \) of the quadratic equation \( x^{2} - 6x - 2 = 0 \). This can be done using the quadratic formula: \( \alpha, \beta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a=1, b=-6, \) and \( c=-2 \). Evaluate this to get the roots.

Step 2: Express Recurrence Relation

Recognize that \( a_n = \alpha^n - \beta^n \) satisfies a recurrence relation derived from the characteristic polynomial of the roots. Use the equation obtained to find \( a_{n+2} - 6a_{n+1} - 2a_n = 0 \).

Step 3: Base Cases

To use the recurrence relation, find the initial values: \( a_0 = \alpha^0 - \beta^0 = 0 \), \( a_1 = \alpha - \beta \), and \( a_2 = \alpha^2 - \beta^2 = (\alpha + \beta)(\alpha - \beta) = 6(\alpha - \beta) \).

Step 4: Use the Recurrence to Find Needed Terms

Apply the recurrence relation to calculate \( a_8, a_9, \text{ and } a_{10} \) using successive terms: \[ a_{n+2} = 6a_{n+1} + 2a_n \]. Compute \( a_8, a_9, \) and \( a_{10} \) this way, starting from \( a_1 \) and \( a_2 \).

Step 5: Compute the Expression

Using the calculated values for \( a_8, a_9, \) and \( a_{10} \), substitute into the given expression: \[ \frac{a_{10} - 2a_8}{2a_9} \]. Evaluate the expression with the found values.

Step 6: Simplify the Expression

Simplify the result from Step 5 to find the numerical value. Ensure that all calculations follow through correctly to yield the final value of the expression.

Key Concepts

Recurrence Relation

A recurrence relation is a way to define a sequence of numbers where each term is defined as a function of its preceding terms. In the given problem, we see a sequence defined by the relation:

• \( a_n = \alpha^n - \beta^n \)

To fully use this setup, we derive a specific form of a recurrence relationship from the properties of the roots of the characteristic polynomial.
The recurrence relation can take form as follows:

• \( a_{n+2} = 6a_{n+1} + 2a_n \)

This equation illustrates that each term in the series can be directly calculated from its two preceding terms.
Such relations are incredibly useful because they allow you to calculate any term within the sequence, given you successfully identify a set of base cases.

Roots of Equation

In the context of quadratic equations, the roots are values of \( x \) that make the equation equal to zero. For the equation \( x^2 - 6x - 2 = 0 \), finding the roots involves using the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Substituting \( a = 1 \), \( b = -6 \), and \( c = -2 \) allows us to calculate:

• \( \alpha, \beta = 3 \pm \sqrt{11} \)

The roots here are crucial since they determine the behavior and characteristics of sequences in which they are integrated.
In problems like these, the roots help in establishing important relationships, such as the recurrence relations discussed previously.

Characteristic Polynomial

The characteristic polynomial is pivotal when working with recurrence relations. It connects directly to the roots of a quadratic equation and serves as a bridge to derive recurrence relations.
For the quadratic \( x^2 - 6x - 2 = 0 \), the characteristic polynomial inherently captures the form:

• \( t^2 - 6t - 2 \)

By constructing a characteristic polynomial, we can formulate a recurrence relation that sequences, like \( a_n = \alpha^n - \beta^n \), must satisfy.
Solving for the characteristic equation's roots can guide you in setting up the sequence's behavior and deriving formulas to find new terms efficiently.