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Answer:

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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.

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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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Keragra
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