x2−x−1=0 roots =α,β
α2−α−1=0⇒αn+1=αn+αn−1β2−β−1=0⇒βn+1=βn+βn−1+pn+1=pn+pn−1
29=pn+11
pn=18
pn2=324
JEE Main 2021 — Mathematics Algebra
Let α and β be two real numbers such that α+β=1 and αβ=−1. Let pn=(α)n+(β)n, pn−1=11 and pn+1=29 for some integer n⩾1. Then, the value of pn2 is______.
Held on 26 Feb 2021 · Verified 6 Jul 2026.
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