# Let $\alpha$ and $\beta$ be two real numbers such that $\alpha+\beta=1$ and $\alpha \beta=-1$.

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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