Let α and β be the roots of the equation x² - x - 1 = 0. If pₖ = (α)ᵏ + (β)ᵏ, k ≥ 1, then which of t

Q: Why can't we just use the quadratic formula? It seems more direct.

While the quadratic formula gives you the exact values of $\alpha$ and $\beta$, it introduces messy radicals like $\frac{1 \pm \sqrt{5}}{2}$. Raising these to the power of 5 is a recipe for calculation errors. Newton's Sums allow you to bypass the roots entirely and work with the structure of the equation itself, which is the hallmark of an elite JEE approach.

Q: How did you derive $P_k = P_{k-1} + P_{k-2}$?

Since $\alpha$ is a root, $\alpha^2 - \alpha - 1 = 0$, which means $\alpha^2 = \alpha + 1$. Multiply by $\alpha^{k-2}$ to get $\alpha^k = \alpha^{k-1} + \alpha^{k-2}$. Doing the same for $\beta$ and adding the two equations gives $\alpha^k + \beta^k = (\alpha^{k-1} + \beta^{k-1}) + (\alpha^{k-2} + \beta^{k-2})$, which is exactly $P_k = P_{k-1} + P_{k-2}$.

Q: Is there a way to check if my $P_k$ values are correct?

Yes! You can always verify $P_2$ using the identity $P_2 = (\alpha + \beta)^2 - 2\alpha\beta$. If your $P_2$ is correct, the rest of the sequence will follow logically from the recurrence relation. Always double-check your $P_1$ and $P_2$ before proceeding to higher powers.

Question

Let $\alpha$ and $\beta$ be the roots of the equation $x^2 - x - 1 = 0$. If $p_k = (\alpha)^k + (\beta)^k, k \ge 1$, then which of the following statements is not true?

Options:

(A) $P_5 = P_2 	imes P_3$

(B) $P_1 + P_2 + P_3 + P_4 + P_5 = 25$

(C) $P_3 = P_5 - P_4$

(D) $P_5 = 11$

Sigma Percentile · Accepted Answer

### The Beauty of Hidden Patterns

Welcome, future engineers! Today, we are going to peel back the layers of a problem that, at first glance, looks like a tedious exercise in exponentiation.

You see the equation $x^2 - x - 1 = 0$ and the sequence $P_k = \alpha^k + \beta^k$, and your instinct might be to reach for the quadratic formula. But wait! Stop!

In the world of JEE Advanced, the brute force path is rarely the intended one. We are here to find the elegance, the hidden symmetry, and the shortcut that separates the masters from the rest.

### Phase 1: The Foundation

Let us start with the basics. We have the quadratic equation $x^2 - x - 1 = 0$. We know that $\alpha$ and $\beta$ are its roots.

From Vieta's formulas, we immediately extract the sum and product of these roots. The sum, $\alpha + \beta$, is given by $-\frac{b}{a}$, which is:

$$-\frac{-1}{1} = 1$$

The product, $\alpha\beta$, is $\frac{c}{a}$, which is:

$$\frac{-1}{1} = -1$$

These two values are our building blocks. They are the DNA of the sequence $P_k$. Do not underestimate them; they are all we need to conquer this problem.

### Phase 2: The Magic of Newton's Sums

Now, here is where the magic happens. Since $\alpha$ is a root of $x^2 - x - 1 = 0$, it must satisfy the equation:

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

If we multiply this entire equation by $\alpha^{k-2}$, we get $\alpha^k = \alpha^{k-1} + \alpha^{k-2}$. The same logic applies to $\beta$:

$$\beta^k = \beta^{k-1} + \beta^{k-2}$$

If we add these two equations together, we get $(\alpha^k + \beta^k) = (\alpha^{k-1} + \beta^{k-1}) + (\alpha^{k-2} + \beta^{k-2})$. Substituting our definition of $P_k$, we arrive at the beautiful recurrence relation:

$$P_k = P_{k-1} + P_{k-2}$$

This is the key to the kingdom! We no longer need to calculate powers of irrational numbers; we just need to add the previous two terms.

### Phase 3: The Calculation

Let us build our sequence. We start with $P_1 = \alpha + \beta = 1$.

Next, we find $P_2 = \alpha^2 + \beta^2$. Using the identity $P_2 = (\alpha + \beta)^2 - 2\alpha\beta$, we substitute our values:

$$P_2 = (1)^2 - 2(-1) = 1 + 2 = 3$$

Now, the recurrence relation takes over:

$$P_3 = P_2 + P_1 = 3 + 1 = 4$$

$$P_4 = P_3 + P_2 = 4 + 3 = 7$$

$$P_5 = P_4 + P_3 = 7 + 4 = 11$$

Look at that! We have generated the sequence $1, 3, 4, 7, 11$ with simple addition. No radicals, no complex algebra, just pure, logical flow.

### Phase 4: The Verification

Now, let us test the options. Option 4 claims $P_5 = 11$, which we have just confirmed is true.

Option 3 claims $P_3 = P_5 - P_4$. Rearranging our recurrence $P_5 = P_4 + P_3$, we see that $P_3 = P_5 - P_4$ is indeed true.

Finally, let us check Option 1: $P_5 = P_2 	imes P_3$. The left side is $11$. The right side is:

$$P_2 	imes P_3 = 3 	imes 4 = 12$$

Since $11 
eq 12$, this statement is false. And there you have it! We have navigated the problem with precision and grace.

Remember, in JEE, the most powerful tool is not the calculator, but your ability to see the underlying structure of the math. Keep practicing, keep questioning, and keep falling in love with the process!

Animated Solution for Mathematics - Quadratic Equations: Let $α$ and $β$ be the roots of the equation $x^{2} - x - 1 = 0$ . If $p_{k} = (α)^{k} + (β)^{k}, k \geq 1$ , then which of the following statements is not true?

Select Answer:

Visualized Solution

Identify the Equation and Roots

Sum and Product of Roots

Derive Newton's Sums Relation

Calculate $P_{1}$ and $P_{2}$

Calculate $P_{3}$ using Recurrence

Calculate $P_{4}$

Calculate $P_{5}$

Verify Option 4 and Option 3

Verify Option 1 (The False Statement)

The Sigma Insight: Relation Between Roots and Coefficients

The Beauty of Hidden Patterns

Phase 1

Phase 2

Phase 3

Phase 4

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