Let α and β be the roots of x2−x−1=0. If Pk=(α)k+(β)k,k≥1 then prove thata) P5=11$b) P1+P2+P3+P4+P5=26

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Erik Cantu

2022-03-23

Let

α and β

be the roots of

x^{2} - x - 1 = 0

. If

P_{k} = {(α)}^{k} + {(β)}^{k}, k \geq 1

then prove that
a)

P_{5} = 11 $

P_{1} + P_{2} + P_{3} + P_{4} + P_{5} = 26

Jesse Gates

Beginner2022-03-24Added 19 answers

x^{2} - x - 1 = 0 \approx\sim (1)

has roots as a, b then

P_{k} = a^{k} + b^{k}, P_{0} = 2, P_{1} = a + b = 1

and

a^{2} - a - 1 = 0 \approx\sim (2), \approx b^{2} - b - 1 = 0 \approx\sim (3)

Multiply Eq.(2) once by

a^{k}

and (3) by

b^{k}

. Adding these two Eqs. you get

(a^{k + 2} + b^{k + 2}) - (a^{k + 1} + b^{k + 1}) - (a^{k} + b^{k}) = 0 \Rightarrow P_{k + 2} = P_{k + 1} + P_{k} \approx\sim (4)

P_{5} = P_{4} + P_{3}, P_{4} = P_{3} + P_{2}, P_{3} = P_{2} + P_{1}, P_{2} = P_{1} + P_{0} = a + b + 2 = 3

\Rightarrow P_{3} = 3 + 1 = 4, \Rightarrow P_{4} = 4 + 3 = 7, P_{5} = 7 + 4 = 11

P_{1} + P_{2} + P_{3} + P_{4} + P_{5} = 1 + 3 + 4 + 7 + 11 = 26

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