Let p , q be integers and let be the roots of the equation x 2 - x - 1 = 0 , where . For n = 0 , 1 , 2 , … , let a n = p n + q n . FACT: If a and b are rational numbers and a + b 5 = 0 , then a = 0 = b . Q. a 12 = [2017]

Question

Let $p, q$ be integers and let $α, β$ be the roots of the equation $x^{2} - x - 1 = 0,$ where $α \neq β$ . For $n = 0, 1, 2, \dots$ , let $a_{n} = p α^{n} + q β^{n} .$

FACT: If $a$ and $b$ are rational numbers and $a + b \sqrt{5} = 0,$ then $a = 0 = b .$ [2017]

Q. $a_{12} =$

Smart Achievers · Accepted Answer

(2)

$α^{2} = α + 1$

$β^{2} = β + 1$

$a_{n} = p α^{n} + q β^{n}$

$= p (α^{n - 1} + α^{n - 2}) + q (β^{n - 1} + β^{n - 2})$

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

$∴$ $a_{12} = a_{11} + a_{10}$

Solution

Q.
Let $p, q$ be integers and let $α, β$ be the roots of the equation $x^{2} - x - 1 = 0,$ where $α \neq β$ . For $n = 0, 1, 2, \dots$ , let $a_{n} = p α^{n} + q β^{n} .$

FACT: If $a$ and $b$ are rational numbers and $a + b \sqrt{5} = 0,$ then $a = 0 = b .$ [2017]

Q. $a_{12} =$

1 $a_{11} - a_{10}$

2 $a_{11} + a_{10}$

3 $2 a_{11} + a_{10}$

4 $a_{11} + 2 a_{10}$

Ans.
(2)

$α^{2} = α + 1$

$β^{2} = β + 1$

$a_{n} = p α^{n} + q β^{n}$

$= p (α^{n - 1} + α^{n - 2}) + q (β^{n - 1} + β^{n - 2})$

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

$∴$ $a_{12} = a_{11} + a_{10}$

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Solution

Q. Let p, q be integers and let α,β be the roots of the equation x2-x-1=0, where α≠β. For n=0,1,2,…, let an=pαn+qβn. FACT: If a and b are rational numbers and a+b5=0, then a=0=b. [2017] Q. a12=

1 a11-a10

2 a11+a10

3 2a11+a10

4 a11+2a10

Ans. (2) α2=α+1 β2=β+1 an=pαn+qβn =p(αn-1+αn-2)+q(βn-1+βn-2) =an-1+an-2 ∴ a12=a11+a10