JEE PYQ: Quadratic Equation Question 14
Question 14 - 2021 (26 Feb Shift 2)
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 $n \geq 1$. Then, the value of $P_n^2$ is
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Answer: (324)
Solution
Given, $\alpha + \beta = 1$, $\alpha\beta = -1$
$\therefore$ Quadratic equation with roots $\alpha, \beta$ is $x^2 - x - 1 = 0 \Rightarrow \alpha^n = \alpha + 1$
Multiplying both sides by $\alpha^{n-1}$: $\alpha^{n+1} = \alpha^n + \alpha^{n-1}$ …(1)
Similarly, $\beta^{n+1} = \beta^n + \beta^{n-1}$ …(2)
Adding (1) & (2): $(\alpha^{n+1} + \beta^{n+1}) = (\alpha^n + \beta^n) + (\alpha^{n-1} + \beta^{n-1})$
$\Rightarrow P_{n+1} = P_n + P_{n-1}$
$= 29 = P_n + 11$ (Given, $P_{n+1} = 29, P_{n-1} = 11$)
$\Rightarrow P_n = 18$
$\therefore P_n^2 = 18^2 = 324$
BETA
