$if α and β are the roots of x^{2} - 2 x + 4 = 0 then α^{n} + β^{n} =$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$2^{n} \cos (n π / 2)$

$2^{n + 1} \cos (nπ / 3)$

$2^{n - 1} \sin (n π / 6)$

$2^{n + 1} \sin (n π / 3)$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{rcl} x^{2} - 2 x + 4 & = & 0 \Rightarrow \Rightarrow x = \frac{2 \pm \sqrt{4 - 16}}{2} \end{array}$

$\begin{array}{rcl} x & = & \frac{2 \pm \sqrt{- 12}}{2} = \frac{2 \pm i 2 \sqrt{3}}{2} = 1 \pm i \sqrt{3} \end{array}$

$\begin{array}{rcl} α^{n} + β^{n} & = & {(1 + i \sqrt{3})}^{n} + {(1 - i \sqrt{3})}^{n} \\ = & 2^{n} [{(\frac{1}{2} + \frac{i \sqrt{3}}{2})}^{n} + {(\frac{1}{2} - \frac{i \sqrt{3}}{2})}^{n}] \\ = & 2^{n} ({(\cos \frac{π}{3} + i \sin \frac{π}{3})}^{n} + {(\cos \frac{π}{3} - i \sin \frac{π}{3})}^{n}) \\ = & 2^{n} (\cos n \frac{π}{3} + i \sin n \frac{π}{3} + \cos n \frac{π}{3} - i \sin n \frac{π}{3}) \\ = & 2^{n} (2 \cos n \frac{π}{3}) = 2^{n + 1} \cos \frac{nπ}{3} \end{array}$