Let α and β be two real numbers such that α+β=1 and α β=-1 . Let p n =(α) n (β) n p n -1=11 and p n +1=29 for some integer n ≥ 1. Then, the value of p n 2 is .

Question

Let α and β be two real numbers such that α+β=1 and α β=-1 . Let p n =(α) n (β) n p n -1=11 and p n +1=29 for some integer n ≥ 1. Then, the value of p n 2 is . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

x 2- x -1=0 roots =α, β α2-α-1=0 ⇒ α n +1=α n +α n -1 β2-β-1=0 ⇒ β n +1=β n +β n -1 (+/ P n +1= P n + P n -1) 29= P n +11 P n =18 P n 2=324

Q. Let $α$ and $β$ be two real numbers such that $α + β = 1$ and $α β = - 1.$ Let $p_{n} = (α)^{n} (β)^{n}$ $p_{n - 1} = 11$ and $p_{n + 1} = 29$ for some integer $n \geq 1$ . Then, the value of $p_{n}^{2}$ is _______.

$x^{2} - x - 1 = 0$ roots $= α, β$
$α^{2} - α - 1 = 0 \Rightarrow α^{n + 1} = α^{n} + α^{n - 1}$
$β^{2} - β - 1 = 0 \Rightarrow β^{n + 1} = β^{n} + β^{n - 1}$
$\frac{+}{P _{n + 1} = P _{n} + P _{n - 1}}$
$29 = P_{n} + 11$
$P_{n} = 18$
$P_{n}^{2} = 324$

Q. Let α and β be two real numbers such that α+β=1 and αβ=−1. Let pn​=(α)n(β)n pn−1​=11 and pn+1​=29 for some integer n≥1. Then, the value of pn2​ is _______.

x2−x−1=0 roots =α,β α2−α−1=0⇒αn+1=αn+αn−1 β2−β−1=0⇒βn+1=βn+βn−1 Pn+1​=Pn​+Pn−1​+​ 29=Pn​+11 Pn​=18 Pn2​=324

Q. Let $α$ and $β$ be two real numbers such that $α + β = 1$ and $α β = - 1.$ Let $p_{n} = (α)^{n} (β)^{n}$ $p_{n - 1} = 11$ and $p_{n + 1} = 29$ for some integer $n \geq 1$ . Then, the value of $p_{n}^{2}$ is _______.

$x^{2} - x - 1 = 0$ roots $= α, β$
$α^{2} - α - 1 = 0 \Rightarrow α^{n + 1} = α^{n} + α^{n - 1}$
$β^{2} - β - 1 = 0 \Rightarrow β^{n + 1} = β^{n} + β^{n - 1}$
$\frac{+}{P _{n + 1} = P _{n} + P _{n - 1}}$
$29 = P_{n} + 11$
$P_{n} = 18$
$P_{n}^{2} = 324$