Let Pn = αⁿ + βⁿ, n ∈ N. If P10 = 123, P9 = 76, P8 = 47 and P1 = 1, then the quadratic equation havi

Q: Wait, why can we assume $P_n = P_{n-1} + P_{n-2}$ just by looking at three numbers?

That is a brilliant question! In competitive exams, when you see a sequence defined by powers like $\alpha^n + \beta^n$, it is almost always linked to a linear recurrence relation. By testing the pattern $P_{10} = P_9 + P_8$, we are forming a hypothesis. Since this relation is consistent with the given $P_1 = 1$ and the structure of the problem, it is the correct path forward. It is not an assumption; it is a deduction based on the mathematical structure of the sequence.

Q: How do I know the characteristic equation is $x^2 - x - 1 = 0$ and not something else?

Great observation. The recurrence relation $P_n = P_{n-1} + P_{n-2}$ can be rewritten as $P_n - P_{n-1} - P_{n-2} = 0$. For any sequence $P_n = \alpha^n + \beta^n$, the characteristic equation is derived from the coefficients of the recurrence. Here, the coefficients are $1$ and $1$ for $P_{n-1}$ and $P_{n-2}$ respectively. Thus, the equation is $x^2 - 1x - 1 = 0$. It is a standard result in the theory of linear recurrence relations!

Q: Is there a faster way to find the equation for $1/\alpha$ and $1/\beta$?

Yes, there is! If a quadratic equation $f(x) = ax^2 + bx + c = 0$ has roots $\alpha$ and $\beta$, then the equation with roots $1/\alpha$ and $1/\beta$ is simply $f(1/x) = 0$. So, if our original equation is $x^2 - x - 1 = 0$, we replace $x$ with $1/x$ to get $(1/x)^2 - (1/x) - 1 = 0$. Multiplying by $x^2$ gives $1 - x - x^2 = 0$, which is equivalent to $x^2 + x - 1 = 0$. This is a powerful shortcut!

Question

Let $P_n = \alpha^n + \beta^n, n \in \mathbb{N}$. If $P_{10} = 123, P_9 = 76, P_8 = 47$ and $P_1 = 1$, then the quadratic equation having roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ is :

Options:

(A) $x^2 - x + 1 = 0$

(B) $x^2 + x - 1 = 0$

(C) $x^2 - x - 1 = 0$

(D) $x^2 + x + 1 = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a sequence of numbers that seem to grow with a life of their own. We are given $P_n = \alpha^n + \beta^n$, with the specific values $P_{10} = 123$, $P_9 = 76$, and $P_8 = 47$.

At first glance, these might look like random integers, but as an elite JEE aspirant, your eyes should be trained to spot the hidden rhythm. If you add $76$ and $47$, you get exactly $123$.

This is not a coincidence; it is the heartbeat of a Fibonacci-style recurrence relation. We have discovered that $P_{10} = P_9 + P_8$, which implies the general rule:
$$P_n = P_{n-1} + P_{n-2}$$

### The Bridge to Algebra

Any sequence of the form $P_n = \alpha^n + \beta^n$ is intimately connected to a quadratic equation. If we have the recurrence $P_n - P_{n-1} - P_{n-2} = 0$, we can map this directly to a characteristic equation.

By treating $P_n$ as $x^2$, $P_{n-1}$ as $x$, and $P_{n-2}$ as $1$, we derive the quadratic equation:
$$x^2 - x - 1 = 0$$

The roots of this equation are precisely $\alpha$ and $\beta$. We can verify this with the given $P_1 = 1$: for our equation $x^2 - x - 1 = 0$, the sum of the roots $\alpha + \beta$ is $-\frac{-1}{1} = 1$. Since $P_1 = \alpha^1 + \beta^1$, this matches perfectly.

### The Transformation

Our goal is to find a new quadratic equation whose roots are $\gamma = \frac{1}{\alpha}$ and $\delta = \frac{1}{\beta}$. We know that any quadratic equation can be written in the form:
$$x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$$

Let us calculate the new sum:
$$\gamma + \delta = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}$$

We know $\alpha + \beta = 1$. From our equation $x^2 - x - 1 = 0$, the product of the roots $\alpha\beta$ is $\frac{c}{a} = \frac{-1}{1} = -1$. Substituting these values, the new sum is:
$$\gamma + \delta = \frac{1}{-1} = -1$$

Next, the new product is:
$$\gamma\delta = \left(\frac{1}{\alpha}ight)\left(\frac{1}{\beta}ight) = \frac{1}{\alpha\beta} = \frac{1}{-1} = -1$$

### The Final Revelation

We have everything we need. The new sum is $-1$ and the new product is $-1$. Plugging these into our standard form $x^2 - (	ext{sum})x + (	ext{product}) = 0$, we get:
$$x^2 - (-1)x + (-1) = 0$$

Simplifying this, we arrive at the final result:
**$$x^2 + x - 1 = 0$$**

It is elegant, it is precise, and it is the solution. You have successfully navigated the relationship between sequences and quadratic equations. Remember, in JEE Advanced, the math is not just about calculation; it is about seeing the underlying structure.

Animated Solution for Mathematics - Quadratic Equations: Let $P_{n} = α^{n} + β^{n}, n \in N$ . If $P_{10} = 123, P_{9} = 76, P_{8} = 47$ and $P_{1} = 1$ , then the quadratic equation having roots $\frac{1}{α}$ and $\frac{1}{β}$ is :

Select Answer:

Visualized Solution

Given Sequence $P_{n}$

Identifying the Pattern

General Recurrence Relation

Characteristic Equation

Verifying with $P_{1}$

Product of Roots

Target Equation Setup

Calculating New Sum of Roots

Calculating New Product of Roots

Forming the Final Equation

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Bridge to Algebra

The Transformation

The Final Revelation

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