JEE Mathematics 2024 Question with Solution

Given: $\alpha$ and $\beta$ are roots of $x^2 - x - 1 = 0$, and $S_n = 2023\alpha^n + 2024\beta^n$.

Find: The correct relation among $S_{10}$, $S_{11}$, and $S_{12}$.

Since $\alpha$ and $\beta$ are roots of $x^2 - x - 1 = 0$, they satisfy

$$\alpha^2 = \alpha + 1, \qquad \beta^2 = \beta + 1$$

Hence, for powers of the roots,

$$\alpha^n = \alpha^{n-1} + \alpha^{n-2}, \qquad \beta^n = \beta^{n-1} + \beta^{n-2}$$

Now,

$$S_n = 2023\alpha^n + 2024\beta^n$$

Substituting the recurrence for $\alpha^n$ and $\beta^n$,

$$S_n = 2023(\alpha^{n-1} + \alpha^{n-2}) + 2024(\beta^{n-1} + \beta^{n-2})$$

So,

$$S_n = (2023\alpha^{n-1} + 2024\beta^{n-1}) + (2023\alpha^{n-2} + 2024\beta^{n-2})$$

Therefore,

$$S_n = S_{n-1} + S_{n-2}$$

Taking $n = 12$,

$$S_{12} = S_{11} + S_{10}$$

Therefore, the correct option is B.

Given: $x^2 - x - 1 = 0$ has roots $\alpha$ and $\beta$.

Find: Which option follows for the sequence $S_n = 2023\alpha^n + 2024\beta^n$.

From the quadratic formula,

$$x = \frac{1 \pm \sqrt{5}}{2}$$

so the roots are

$$\alpha = \frac{1 + \sqrt{5}}{2}, \qquad \beta = \frac{1 - \sqrt{5}}{2}$$

Because each root satisfies the equation,

$$\alpha^2 - \alpha - 1 = 0, \qquad \beta^2 - \beta - 1 = 0$$

which gives

$$\alpha^2 = \alpha + 1, \qquad \beta^2 = \beta + 1$$

Multiplying appropriately by powers, we get the recurrence

$$\alpha^n = \alpha^{n-1} + \alpha^{n-2}, \qquad \beta^n = \beta^{n-1} + \beta^{n-2}$$

Therefore,

$$\begin{aligned} S_n &= 2023\alpha^n + 2024\beta^n \\ &= 2023(\alpha^{n-1} + \alpha^{n-2}) + 2024(\beta^{n-1} + \beta^{n-2}) \\ &= S_{n-1} + S_{n-2} \end{aligned}$$

Thus,

$$S_{12} = S_{11} + S_{10}$$

the solution contains contradictory statements mentioning another relation, but the derived recurrence and the marked correct option both support Option B. Hence, the correct answer is B.