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Q.

Let p,q be integers and let α,β be the roots of the equation, x2−x−1=0, where α≠β .For n=0,1,2,… ,  let an=pαn+qβn .  FACT: If a and b are rational numbers and a+b5=0, then a=0=b If a4=28, then p+2q=a12=

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a

12

b

21

c

14

d

7

e

a11+2a10

f

a11+a10

g

a11−a10

h

2a11+a10

answer is , .

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Detailed Solution

α+β=1 α=1+52,β=1−52αβ=−1 α2=α+1α4=3α+2,β4=3β+2α4=Pα4+qβ428=P(3α+2)+q(3β+2)=8=12(p+q)+352(p−q)⇒p=q=4 ∴p+2q=12α5=5α+βα6=8α+5⇒α12=144α+89;β12=144β+89a12=pα12+qβ12a12==144(pα+qβ)+89(p+q)∴a11+a10=a12
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