Let α, β (α β) be the roots of the quadratic equation x² - x - 4 = 0. If Pₙ = αⁿ - βⁿ, n ∈ N, then

Q: Wait, why can't we just find the roots $\alpha$ and $\beta$ using the quadratic formula and plug them into $P_n = \alpha^n - \beta^n$?

While theoretically possible, it is a trap! The roots are $\frac{1 \pm \sqrt{17}}{2}$. Calculating $\alpha^{16}$ or $\beta^{16}$ would be a nightmare of binomial expansions. The beauty of this problem lies in using the recurrence relation, which bypasses the need to know the actual values of the roots.

Q: How did you know to group the numerator terms like that?

That is a classic pattern recognition skill in JEE algebra. When you see an expression with four terms like $P_{15}P_{16} - P_{14}P_{16} - P_{15}^2 + P_{14}P_{15}$, it is almost always designed to be factored by grouping. Look for common factors in pairs—here, $P_{16}$ and $P_{15}$—and you will see the common binomial factor $(P_{15} - P_{14})$ emerge.

Q: Is the recurrence relation $P_n - P_{n-1} = 4P_{n-2}$ always true for this type of problem?

Yes! For any quadratic equation $x^2 - ax - b = 0$ with roots $\alpha, \beta$, the sequence $P_n = \alpha^n - \beta^n$ will always satisfy the recurrence $P_n = aP_{n-1} + bP_{n-2}$. In our case, $a=1$ and $b=4$, so the relation $P_n = P_{n-1} + 4P_{n-2}$ is a direct consequence of the equation's structure.

Question

Let $\alpha, \beta$ ($\alpha > \beta$) be the roots of the quadratic equation $x^2 - x - 4 = 0$. If $P_n = \alpha^n - \beta^n, n \in \mathbb{N}$, then $\frac{P_{15}P_{16} - P_{14}P_{16} - P_{15}^2 + P_{14}P_{15}}{P_{13}P_{14}}$ is equal to ______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a daunting mountain of algebra. The problem asks you to evaluate a complex expression involving $P_n = \alpha^n - \beta^n$, where $\alpha$ and $\beta$ are roots of $x^2 - x - 4 = 0$.

At first glance, the subscripts $13, 14, 15,$ and $16$ might make you want to reach for a calculator or despair at the sheer volume of calculation. But stop. In the world of JEE Advanced, complexity is often a mask for a deeper, more elegant simplicity.

We are not here to calculate; we are here to uncover the hidden structure.

### The DNA of the Equation

Every quadratic equation has a secret, a piece of DNA that defines its roots. If $\alpha$ is a root of $x^2 - x - 4 = 0$, then by definition, it must satisfy the equation:

$$\alpha^2 - \alpha - 4 = 0$$

This means $\alpha^2 = \alpha + 4$. The same is true for $\beta$. This is our foundation.

We don't need the values of $\alpha$ and $\beta$; we only need to know how they behave. By multiplying this relation by $\alpha^{n-2}$, we get:

$$\alpha^n = \alpha^{n-1} + 4\alpha^{n-2}$$

Similarly, $\beta^n = \beta^{n-1} + 4\beta^{n-2}$. This is the bridge between the powers.

### The Birth of the Recurrence Relation

Now, let's subtract the $\beta$ equation from the $\alpha$ equation:

$$(\alpha^n - \beta^n) = (\alpha^{n-1} - \beta^{n-1}) + 4(\alpha^{n-2} - \beta^{n-2})$$

Look closely. Does this look familiar? It is exactly the definition of $P_n$!

Substituting $P_n$, we get the master recurrence relation:

$$P_n = P_{n-1} + 4P_{n-2}$$

Or, more usefully, $P_n - P_{n-1} = 4P_{n-2}$. This simple equation is the key that will unlock the entire problem. It tells us that the difference between consecutive terms is just a scaled version of the term two steps back.

### The Algebraic Dance

Now, let's turn our attention to the numerator: $P_{15}P_{16} - P_{14}P_{16} - P_{15}^2 + P_{14}P_{15}$. It looks like a tangled mess, but let's apply the art of factoring by grouping.

Group the first two terms and the last two terms:

$$P_{16}(P_{15} - P_{14}) - P_{15}(P_{15} - P_{14})$$

Suddenly, the common binomial $(P_{15} - P_{14})$ appears. Factoring it out, we get:

$$(P_{15} - P_{14})(P_{16} - P_{15})$$

The complexity has vanished, replaced by two simple, elegant factors.

### The Grand Finale

We are in the endgame. We have our factored numerator: $(P_{15} - P_{14})(P_{16} - P_{15})$. We have our denominator: $P_{13}P_{14}$.

Using our master key, $P_n - P_{n-1} = 4P_{n-2}$, we can transform the numerator:

For $n=15$, $P_{15} - P_{14} = 4P_{13}$.

For $n=16$, $P_{16} - P_{15} = 4P_{14}$.

The numerator becomes $(4P_{13})(4P_{14})$. The fraction is now:

$$\frac{4P_{13} 	imes 4P_{14}}{P_{13}P_{14}}$$

The $P_{13}$ and $P_{14}$ terms cancel out perfectly, leaving us with $4 	imes 4 = 16$.

We have conquered the mountain, not by brute force, but by understanding the beautiful, underlying symmetry of the equation. The final answer is **16**.

Enter Numerical Value:

Visualized Solution

The Quadratic Equation

Roots Satisfy the Equation

The Power Multiplication Trick

Generating Higher Powers

Forming the Recurrence Relation

Rearranging the Relation

Analyzing the Numerator

Factoring by Grouping

Applying the Recurrence Relation

Final Substitution and Evaluation

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The DNA of the Equation

The Birth of the Recurrence Relation

The Algebraic Dance

The Grand Finale

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