2014/097) If $x^2-2x +4$ has roots a and b then show that $a^n + b^n = 2^{n+1} \cos\dfrac{\pi}{3}$ for n positive integer

We have as a and b are roots

$a+ b= 2$ and $ab = 4$

so $a +b = 2 = 2^2 \cos \dfrac{\pi}{3}$ true for 1

$(a^2+b^2) = (a+b)^2 – 2ba = 4 – 8 = -4 = 8 * \cos \dfrac{2\pi}{3}$ true for 2

Let it be true for n =k

We shall prove it by induction

$(a^k+b^k)(a+b) = a^{k+1} +b^{k+1} + ab(a^{k-1} +b^{k-1})$

Or 

$a^{k+1}+b^{k+1}= (a^k+b^k)(a+b)-ab(a^{k-1}+b^{k-1})$

we have

$a^{k+1}+b^{k+1}= (a^k+b^k)(a+b)-ab(a^{k-1}+b^{k-1})$

= $2^{n+1} \cos \dfrac{n\pi}{3} 2^2 \cos \dfrac{\pi}{3} – 4(2^n \cos\dfrac{(n-1)\pi}{3})$

= $2^{n+2}( 2\cos \dfrac{n\pi}{3} \cos \dfrac{\pi}{3}) – 4(2^n \cos\dfrac{(n-1)\pi}{3})$

= $2^{n+2}( \cos \dfrac{(n+1)\pi}{3} + \cos \dfrac{(n-1)\pi}{3}) – 2^{n+2} \cos\dfrac{(n-1)\pi}{3}$
= $2^{n+2}\cos \dfrac{(n+1)\pi}{3} +2^{n+2} \cos \dfrac{(n-1)\pi}{3} – 2^{n+2} \cos\dfrac{(n-1)\pi}{3}$

=  $2^{n+2}\cos \dfrac{(n+1)\pi}{3}$

We have proved the step for induction

Hence proved