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JEE Main 202126 FebruaryEvening Shift
Exam
JEE Main
Year
2021
Date
26 February
Shift
Evening Shift
Subject
Mathematics
Chapter
Quadratic Equation

JEE Main 2021 · 26 February · Evening Shift · Mathematics · Quadratic Equation

Let α and β be two real numbers such that α + β = 1 and αβ = - 1 ....

Let α\alpha and β\beta be two real numbers such that α+β=1\alpha + \beta = 1 and αβ=−1\alpha\beta = - 1. Let pn=(α)n+(β)n{p}_{n} = {\left(\alpha\right)}^{n} + {\left(\beta\right)}^{n}, pn−1=11{p}_{n - 1} = 11 and pn+1=29{p}_{n + 1} = 29 for some integer n⩾1n ⩾ 1. Then, the value of pn2{p}_{n}^{2} is______.

Correct answer

0

Step-by-step solution

x2−x−1=0{x}^{2} - x - 1 = 0 roots =α,β= \alpha, \beta

α2−α−1=0⇒αn+1=αn+αn−1β2−β−1=0⇒βn+1=βn+βn−1+pn+1=pn+pn−1{\alpha}^{2} - \alpha - 1 = 0 ⇒ {\alpha}^{n + 1} = {\alpha}^{n} + {\alpha}^{n - 1}{\beta}^{2} - \beta - 1 = 0 ⇒ {\beta}^{n + 1} = {\beta}^{n} + {\beta}^{n - 1} + {p}_{n + 1} = {p}_{n} + {p}_{n - 1}

29=pn+1129 = {p}_{n} + 11

pn=18{p}_{n} = 18

pn2=324{p}_{n}^{2} = 324

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