If α,β are the roots of x2−2x+4=0 and for any n∈N, αn+βn=k cosnπ3 then k=

Given Equation x2−2x+4=0x=2±4−4(1)42=2±2i32=1±i3∴ α=1+i3,β=1−i3α=2(12+i32)=2(cosπ3+isinπ3)∴ β=2(cosπ3−isinπ3)∴ αn+βn=2n(cosπ3+isinπ3)n+2n(cosπ3−isinπ3)n=2n(cosnπ3+isinnπ3+cosnπ3−isinnπ3)=2n.2 cosnπ3=2n+1cosnπ3∴ K= 2n+1

If α,β are the roots of x2−2x+4=0 and for any n∈N, αn+βn=k cosnπ3 then k=

Given Equation x2−2x+4=0x=2±4−4(1)42=2±2i32=1±i3∴ α=1+i3,β=1−i3α=2(12+i32)=2(cosπ3+isinπ3)∴ β=2(cosπ3−isinπ3)∴ αn+βn=2n(cosπ3+isinπ3)n+2n(cosπ3−isinπ3)n=2n(cosnπ3+isinnπ3+cosnπ3−isinnπ3)=2n.2 cosnπ3=2n+1cosnπ3∴ K= 2n+1

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$I f α, β a r e t h e r o o t s o f x^{2} - 2 x + 4 = 0 a n d f o r a n y n \in N, α^{n} + β^{n} = k \cos \frac{n π}{3} t h e n k =$

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$2^{n} - 1$

$2^{n}$

$2^{n} + 1$

$2^{n + 1}$

answer is B.

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Detailed Solution

$\begin{array}{l} G i v e n E q u a t i o n x^{2} - 2 x + 4 = 0 \\ x = \frac{2 \pm \sqrt{4 - 4 (1) 4}}{2} = \frac{2 \pm 2 i \sqrt{3}}{2} = 1 \pm i \sqrt{3} \\ ∴ α = 1 + i \sqrt{3}, β = 1 - i \sqrt{3} \\ α = 2 (\frac{1}{2} + i \frac{\sqrt{3}}{2}) = 2 (\cos \frac{π}{3} + i \sin \frac{π}{3}) \\ ∴ β = 2 (\cos \frac{π}{3} - i \sin \frac{π}{3}) \\ ∴ α^{n} + β^{n} = 2^{n} {(\cos \frac{π}{3} + i \sin \frac{π}{3})}^{n} + 2^{n} {(\cos \frac{π}{3} - i \sin \frac{π}{3})}^{n} \\ = 2^{n} (\cos \frac{n π}{3} + i \sin \frac{n π}{3} + \cos \frac{n π}{3} - i \sin \frac{n π}{3}) \\ = 2^{n} .2 \cos \frac{n π}{3} = 2^{n + 1} \cos \frac{n π}{3} \\ ∴ K = 2^{n + 1} \end{array}$