Let α, β; α β, be the roots of the equation x² - √2x - √3 = 0. Let P_n = αⁿ - βⁿ, n ∈ N. Then (11√

Q: Why can't I just find the roots? The quadratic formula is standard!

The trap is in the complexity. Calculating $\alpha = \frac{\sqrt{2} + \sqrt{2 + 4\sqrt{3}}}{2}$ and then raising it to the 12th power is a recipe for disaster. Newton's Sums bypasses the roots entirely, focusing on the relationship between coefficients and powers. It is the elegant path, not the brute-force path.

Q: How do I know the recurrence relation is $P_{n+2} - \sqrt{2}P_{n+1} - \sqrt{3}P_n = 0$?

The recurrence is derived directly from the characteristic equation. Since $\alpha$ and $\beta$ are roots, they satisfy $\alpha^2 - \sqrt{2}\alpha - \sqrt{3} = 0$, which means $\alpha^2 = \sqrt{2}\alpha + \sqrt{3}$. Multiplying by $\alpha^n$, we get $\alpha^{n+2} = \sqrt{2}\alpha^{n+1} + \sqrt{3}\alpha^n$. Subtracting the corresponding equation for $\beta$ yields the relation for $P_n$.

Q: Is there a shortcut to the final simplification?

Yes, look for the symmetry. The coefficients in the expression $(11\sqrt{3} - 10\sqrt{2})P_{10} + (11\sqrt{2} + 10)P_{11} - 11P_{12}$ are specifically designed to cancel out when you substitute $P_{12}$. Trust the algebra; it is designed to collapse into a much simpler form.

Question

Let $\alpha, \beta; \alpha > \beta$, be the roots of the equation $x^2 - \sqrt{2}x - \sqrt{3} = 0$. Let $P_n = \alpha^n - \beta^n, n \in \mathbb{N}$. Then $(11\sqrt{3} - 10\sqrt{2})P_{10} + (11\sqrt{2} + 10)P_{11} - 11P_{12}$ is equal to

Options:

(A) $10\sqrt{3}P_9$

(B) $11\sqrt{3}P_9$

(C) $10\sqrt{2}P_9$

(D) $11\sqrt{2}P_9$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, student. Today, we are going to dismantle a problem that, at first glance, looks like a nightmare of high-power algebra. You see an expression involving $P_{10}$, $P_{11}$, and $P_{12}$, where $P_n = \alpha^n - \beta^n$.

Your instinct might be to reach for the quadratic formula, to find the exact values of $\alpha$ and $\beta$, and then to start calculating powers. Stop. Take a deep breath.

In the world of JEE Advanced, brute force is rarely the intended path. When you see powers of roots, you are not looking at a calculation problem; you are looking at a structural problem. We are going to use the elegance of **Newton's Sums** to solve this without ever knowing the exact values of $\alpha$ or $\beta$.

### The Recurrence Relation

Our quadratic equation is $x^2 - \sqrt{2}x - \sqrt{3} = 0$. Since $\alpha$ and $\beta$ are roots, they must satisfy this equation.

This means $\alpha^2 = \sqrt{2}\alpha + \sqrt{3}$ and $\beta^2 = \sqrt{2}\beta + \sqrt{3}$. If we multiply the first by $\alpha^n$ and the second by $\beta^n$, we get:

$$\alpha^{n+2} = \sqrt{2}\alpha^{n+1} + \sqrt{3}\alpha^n$$

$$\beta^{n+2} = \sqrt{2}\beta^{n+1} + \sqrt{3}\beta^n$$

Subtracting these two equations gives us the recurrence relation:

$$P_{n+2} - \sqrt{2}P_{n+1} - \sqrt{3}P_n = 0$$

This is our golden key. It tells us that any term in our sequence is just a linear combination of the two preceding terms. We don't need to calculate $\alpha^{12}$; we only need to know how it relates to $P_{11}$ and $P_{10}$.

### The Algebraic Dance

Let us look at the expression we are tasked to evaluate:

$$E = (11\sqrt{3} - 10\sqrt{2})P_{10} + (11\sqrt{2} + 10)P_{11} - 11P_{12}$$

This looks intimidating, but notice the $P_{12}$ term. Using our recurrence relation, we know that $P_{12} = \sqrt{2}P_{11} + \sqrt{3}P_{10}$.

Let us substitute this into our expression $E$. This is where the magic happens:

$$E = (11\sqrt{3} - 10\sqrt{2})P_{10} + (11\sqrt{2} + 10)P_{11} - 11(\sqrt{2}P_{11} + \sqrt{3}P_{10})$$

### The Great Cancellation

Now, let us expand this carefully. Distributing the $-11$ gives us $-11\sqrt{2}P_{11} - 11\sqrt{3}P_{10}$.

Now, group the terms by $P_{10}$ and $P_{11}$:

For $P_{10}$, we have $(11\sqrt{3} - 10\sqrt{2} - 11\sqrt{3})P_{10}$. The $11\sqrt{3}$ and $-11\sqrt{3}$ cancel out perfectly, leaving us with $-10\sqrt{2}P_{10}$.

For $P_{11}$, we have $(11\sqrt{2} + 10 - 11\sqrt{2})P_{11}$. Again, the $11\sqrt{2}$ and $-11\sqrt{2}$ vanish, leaving us with $10P_{11}$.

Our monstrous expression has collapsed into $10P_{11} - 10\sqrt{2}P_{10}$.

### The Final Reveal

We are almost there. Factor out the $10$ to get $10(P_{11} - \sqrt{2}P_{10})$.

Does this look familiar? Go back to our recurrence relation, but this time set $n=9$. We get:

$$P_{11} - \sqrt{2}P_{10} - \sqrt{3}P_9 = 0$$

This implies that $P_{11} - \sqrt{2}P_{10} = \sqrt{3}P_9$. Substitute this back into our simplified expression, and we get:

$$10(\sqrt{3}P_9) = 10\sqrt{3}P_9$$

We have arrived at the answer. It is not just about getting the right option; it is about seeing the symmetry, trusting the recurrence, and watching the complexity dissolve into simplicity. That is the true joy of mathematics.

Animated Solution for Mathematics - Quadratic Equations: Let $α, β; α > β$ , be the roots of the equation $x^{2} - 2 x - 3 = 0$ . Let $P_{n} = α^{n} - β^{n}, n \in N$ . Then $(113 - 102) P_{10} + (112 + 10) P_{11} - 11 P_{12}$ is equal to

Select Answer:

Visualized Solution

Analyze the Quadratic Equation

Apply Newton's Sums Relation

Express $P_{12}$ in Lower Terms

Substitute into the Expression

Simplify $P_{10}$ Coefficients

Simplify $P_{11}$ Coefficients

Factor and Use Recurrence Again

Final Result

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Recurrence Relation

The Algebraic Dance

The Great Cancellation

The Final Reveal

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