a+b+c+d=2, a^2 +b^2 +c^2 +d^2 =2 a^3 +b^3 +c^3 +d^3 =−4, a^4 +b^4 +c^4 +d^4 =−6 find real value of a^(2023) +b^(2023) +c^(2023) +d^(2023) .


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Question Number 212320 by RojaTaniya last updated on 10/Oct/24

$a + b + c + d = 2, a^{2} + b^{2} + c^{2} + d^{2} = 2$ $a^{3} + b^{3} + c^{3} + d^{3} = - 4, a^{4} + b^{4} + c^{4} + d^{4} = - 6$ $f i n d r e a l v a l u e o f a^{2023} + b^{2023} + c^{2023} + d^{2023} .$
Commented by mr W last updated on 10/Oct/24

$= 2^{1012}$
	Answered by mr W last updated on 10/Oct/24

	$p_{1} = e_{1} = 2$ $p_{2} = e_{1} p_{1} - 2 e_{2} = 2 \Rightarrow e_{2} = 1$ $p_{3} = e_{1} p_{2} - e_{2} p_{1} + 3 e_{3} = - 4 \Rightarrow e_{3} = - 2$ $p_{4} = e_{1} p_{3} - e_{2} p_{2} + e_{3} p_{1} - 4 e_{4} = - 6 \Rightarrow e_{4} = - 2$ $p_{n} = r_{1}^{n} + r_{2}^{n} + r_{3}^{n} + r_{4}^{n}$ $r_{1 - 4} a r e r o o t s o f$ $x^{4} - 2 x^{3} + x^{2} + 2 x - 2 = 0$ $(x^{2} - 1) (x^{2} - 2 x + 2) = 0$ $r_{1, 2} = \pm 1, r_{3, 4} = 1 \pm i = \sqrt{2} (\cos \frac{π}{4} \pm i \sin \frac{π}{4})$ $\Rightarrow p_{n} = 1 + {(- 1)}^{n} + 2^{\frac{n}{2} + 1} \cos \frac{n π}{4}$ $\Rightarrow p_{2023} = 1 + {(- 1)}^{2023} + 2^{\frac{2023}{2} + 1} \cos \frac{2023 π}{4}$ $= 1 - 1 + 2^{1012} \sqrt{2} \cos \frac{π}{4} = 2^{1012} ✓$
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