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Let \( \alpha \) and \( \beta \) be two real numbers such that \( \alpha+\beta=1 \) and \( \alpha \beta=-1 \). Let \( p _{ n }=(\alpha)^{ n }+(\beta)^{ n } \), \( p _{ n -1}=11 \) and \( p _{ n +1}=29 \) for some integer \( n \geqslant 1 \). Then, the value of \( p _{ n }^{2} \) is

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\(\alpha + \beta = 1\)----(1)

and αβ = -1, α and β are real numbers.

\(\because\) (α + ß)2 = (α + ß)2 - 4 α ß

 = 12 - 4 x -1

 =  1 + 4 = 5

\(\therefore\) α - ß = \(\pm\sqrt5\)

α - ß = -√5 or α - ß = √5----(2)

\(\therefore\) 2α = 1 - √5 or 2α = 1 + √5 (from (1) and (2))

\(\therefore\) α = \(\frac{1-\sqrt5}2\) or α = \(\frac{1+\sqrt5}2\)

⇒ 1 - α  = \(\frac{1+\sqrt5}2\) or ß  = \(\frac{1-\sqrt5}2\) 

Let α = \(\frac{1+\sqrt5}2\) and   ß  = \(\frac{1-\sqrt5}2\) 

Pn = (\(\frac{1+\sqrt5}2\))n + (\(\frac{1-\sqrt5}2\))

 = \(\frac1{2^n}((1+\sqrt5)^n+(1-\sqrt5)^n)\)----(3)

P1 = \(\frac12\)((1 + √5) + (1 - √5)) = 2/2 = 1

P2 = \(\frac14\)((1 + √5)2 + (1 - √5)2)

 = \(\frac14\)(1 + 5 + 2√5+1 + 5 - 2√5)

 = 12/4 = 3

P3 = \(\frac18\)((1 + √5)3 + (1 - √5)3)

 = \(\frac18\)((1 + 3√5 + 3 x 5 + 5√5) + (1 - 3√5 + 3 x 5 - 5√5))

 = \(\frac18\) (1 + 15 + 1 + 15) = 32/8 = 4

P4 = \(\frac{1}{16}\)((1 + 4√5 + 6 x 5 + 4 x 5√5 + 25) + (1 - 4√5 + 6 x 5 - 4 x 5 x 5√5 + 25))

 = \(\frac1{16}\)(2(1 + 30 + 25)) = 56/8 = 7

P5 = \(\frac1{32}\) ((1 + 5√5 + 10 x 5 + 10 x 5√5 + 5 x 25 + 25√5) - (1 - 5√5 + 10 x 5 - 10 x 5√5 + 5 x 25 - 25√5))

 = \(\frac1{32}\) (2 x (1 + 50 + 125)) = 176/16 = 11

\(\therefore\) n - 1 = 5

n = 6

P4 = P6 = \(\frac1{64}\)(2(1 + 15 x 5 + 15 x 25 + 125))

 = 1/32 x 576 = 18

\(\therefore\) Pn2 = 182 = 324

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