Let α and β be two real numbers such that α + β = 1 and αβ = -1. Let Pₙ = (α)ⁿ + (β)ⁿ, Pₙ₋₁ = 11 and

Q: Wait, why can't we just solve for $\alpha$ and $\beta$ using the quadratic formula?

You absolutely could! However, the roots of $x^2 - x - 1 = 0$ are $\frac{1 \pm \sqrt{5}}{2}$, which are irrational. Raising these to the power of $n$ and simplifying would be an algebraic nightmare. The beauty of this problem lies in using the recurrence relation, which bypasses the need to calculate the roots explicitly.

Q: How did you know to multiply by $\alpha^{n-1}$?

This is a classic technique in sequence problems involving roots. We have $\alpha^2 = \alpha + 1$. To get $\alpha^n$ and $\alpha^{n+1}$ involved, we multiply by $\alpha^{n-1}$. This shifts the powers perfectly to match the definition of $P_n$ and $P_{n+1}$.

Q: Is this recurrence relation always true for any quadratic?

Yes! This is a specific case of Newton's Sums. For any quadratic equation $x^2 - ax - b = 0$, the sum of the $n$-th powers of the roots $P_n$ will always satisfy the recurrence $P_{n+1} = aP_n + bP_{n-1}$.

Question

Let $\alpha$ and $\beta$ be two real numbers such that $\alpha + \beta = 1$ and $\alpha\beta = -1$. Let $P_n = (\alpha)^n + (\beta)^n, P_{n-1} = 11$ and $P_{n+1} = 29$ for some integer $n \ge 1$. Then, the value of $P_n^2$ is

Sigma Percentile · Accepted Answer

### The Hidden Elegance of Roots and Sequences

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to peel back the layers of a problem that, at first glance, might look like a tedious exercise in exponentiation.

We are given $\alpha + \beta = 1$ and $\alpha\beta = -1$, and we are asked to find $P_n^2$ where $P_n = \alpha^n + \beta^n$. Many students see this and immediately reach for the quadratic formula, preparing to battle with $\frac{1 \pm \sqrt{5}}{2}$.

But stop! There is a more elegant, more powerful way to see this.

### Phase 1: The Quadratic Foundation

Think of $\alpha$ and $\beta$ not as isolated numbers, but as the DNA of a quadratic equation. If we know the sum of the roots is $1$ and the product is $-1$, we can immediately write down the equation they satisfy:

$$x^2 - (\alpha + \beta)x + \alpha\beta = 0$$

Substituting our known values, we get:

$$x^2 - x - 1 = 0$$

This is the heartbeat of our problem. Because $\alpha$ and $\beta$ are roots, they must obey this equation.

This means:

$$\alpha^2 - \alpha - 1 = 0 \Rightarrow \alpha^2 = \alpha + 1$$

And similarly:

$$\beta^2 - \beta - 1 = 0 \Rightarrow \beta^2 = \beta + 1$$

This simple rearrangement is our key to the kingdom.

### Phase 2: The Power of Recurrence

Now, we want to connect this to our sequence $P_n = \alpha^n + \beta^n$. How do we bridge the gap between $\alpha^2$ and $\alpha^n$? We multiply!

If we take our equation $\alpha^2 = \alpha + 1$ and multiply both sides by $\alpha^{n-1}$, we get:

$$\alpha^2 \cdot \alpha^{n-1} = (\alpha + 1) \cdot \alpha^{n-1}$$

Which simplifies to:

$$\alpha^{n+1} = \alpha^n + \alpha^{n-1}$$

The same logic applies to $\beta$:

$$\beta^{n+1} = \beta^n + \beta^{n-1}$$

Now, watch the magic happen. If we add these two equations together, we get:

$$(\alpha^{n+1} + \beta^{n+1}) = (\alpha^n + \beta^n) + (\alpha^{n-1} + \beta^{n-1})$$

This is exactly the recurrence relation:

$$P_{n+1} = P_n + P_{n-1}$$

This is a beautiful result, often referred to as **Newton's Sums**. It tells us that each term in our sequence is simply the sum of the two preceding terms—just like the Fibonacci sequence!

### Phase 3: The Final Calculation

We are now in the home stretch. The problem provides us with $P_{n-1} = 11$ and $P_{n+1} = 29$.

Substituting these into our recurrence relation:

$$29 = P_n + 11$$

Solving for $P_n$ is straightforward:

$$P_n = 29 - 11 = 18$$

But wait! Do not let your guard down now. The question asks for $P_n^2$. We have found $P_n = 18$, so we must calculate:

$$P_n^2 = 18^2 = 324$$

And there it is. Through the power of recurrence and a deep understanding of how roots behave, we have bypassed the messy irrational numbers and arrived at the solution with absolute precision.

Keep this perspective in your toolkit—whenever you see sums of powers of roots, look for the recurrence. It is the **secret weapon of the JEE topper**.

Animated Solution for Mathematics - Quadratic Equations: Let $α$ and $β$ be two real numbers such that $α + β = 1$ and $α β = - 1$ . Let $P_{n} = (α)^{n} + (β)^{n}, P_{n - 1} = 11$ and $P_{n + 1} = 29$ for some integer $n \geq 1$ . Then, the value of $P_{n}^{2}$ is

Enter Numerical Value:

Visualized Solution

Given Parameters

Forming the Quadratic Equation

Roots Satisfy the Equation

Recurrence for $α$

Recurrence for $β$

Summing to find $P_{n}$

Substituting Given Values

Solving for $P_{n}$

Calculating $P_{n}^{2}$

The Sigma Insight: Relation Between Roots and Coefficients

The Hidden Elegance of Roots and Sequences

Phase 1

Phase 2

Phase 3

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Enter Numerical Value:

Visualized Solution

Given Parameters

Forming the Quadratic Equation

Roots Satisfy the Equation

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Recurrence for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

Summing to find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Pn​

Substituting Given Values

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