Let a and b be two real numbers such that a + b = 1 and ab = -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……

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alok kumar singh | Contributor-Level 10

2 weeks ago

Given $P_{n} = α^{n} + β^{n}, P_{n - 1} = 11 & P_{n + 1} = 29$

$P_{n} = α^{n - 2} . α^{2} + β^{n - 2} . β^{2} . . . . . . . . . (i)$

Now quadratic equation having roots a & b will be x² – (a + b)x + ab = 0

i.e. x² – x – 1 = 0, put x = a and put x = b

So a² = a + 1 & b² = b + 1

(i) $P_{n} = α^{n - 2} (α + 1) + β^{n - 2} (β + 1)$

$P_{n} = P_{n - 1} + P_{n - 2}$

-> ${P_{n}}^{2} = 234$

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Let a and b be two real numbers such that a + b = 1 and ab = -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……

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

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Let a and b be two real numbers such that a + b = 1 and ab = -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……