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Question

Let α and β be two real numbers such that α+β=1and αβ=1. LetPn=(α)n+(β)n,Pn-1=11 and Pn+1=29 for some integer n1Then, the value of Pn2 is______________.


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Solution

Step 1: Finding αn+1 and βn+1

Sum of roots of x2x1=0...(i)

α+β=1

Product of roots

ɑβ=1

Since α be the root of i

α2α1=0α2=α+1α2=α1+α0α1+1=α1+α1-1αn+1=αn+αn-1...(ii)

Similarly βn+1=βn+βn-1...(iii)

Step 2: Finding the value of Pn2

Adding equations (ii)and (iii)

αn+1+βn+1=(α)n+(β)n+[(α)n+1+(β)n+1]Pn+1=Pn+Pn-1Pn=(α)n+(β)n29=Pn+11Pn-1=11&Pn+1=29Pn=29-11Pn=18Pn2=324

Hence, the value of Pn2=324.


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