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Answers
Solution :
Given equation : x² - x - 1 = 0
The two roots of this equation are alpha and beta .
Let us define ,
f(x) = x² - x - 1 ( = 0 )
As alpha and beta are the roots , f( alpha ) and f(beta ) will hold the similar conditions
That is ;
f( alpha ) = alpha² - alpha - 1 = 0
=> alpha ² = alpha + 1
This is the crucial step in solving this .
Do the same for f( beta ) .
We will get ;
=> beta ² = beta + 1.
Now , we have this given that , for n = ( 0,1, 2, ... 12),
a_n = p ( alpha )^n + q( beta)^n
Let us try plugging some values of n here to see what equations we obtain
a_0 = p + q
a_1 = p alpha + q beta
a_2 = p alpha ² + p beta ²'
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Let us now observe these properties .
Simplifying a_2 first
p alpha ^2 + q beta ^2 .
=> p ( alpha + 1) + q ( beta + 1 )
=> p alpha + p beta + p + q
So ,
We can observe that there is a commutating pattern here
a_2 = a_0 + a_1
a_3 = a_2 + a_1
and so on.
That is ,
a_n = a_n-1 + a_n-2
So , a_12 = a_11 + a_10 .
Now , is is given a_4 = 28
=> a_3 + a_2 = 28
=> ( a_2 + a_1) + a_2 = 28
=> 2 a_2 + a_1 = 28
=> 2 ( a_1 + a_0) + a_1 = 28
=> 3 a_1 + a_0 = 28
=> 3 ( p alpha + q beta ) + p + q = 28
=> 3p alpha + 3q beta + p + q = 28
=> p ( alpha + 1) + 2q ( 3/2 beta + 1/2) = 28
From here , we will get p + 2q = 12
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(i) If a4 = 28, then p+ 2q =
(A) 14
(B) 12
(C) 7
(D) 21
(ii) a¹²
(A)a¹¹ – a¹⁰
(B)a¹¹ + a¹⁰
(C) a¹¹ + 2a¹⁰
(D)2a¹¹ + a¹⁰
