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Question Number 95767 by PengagumRahasiamu last updated on 27/May/20

	Answered by prakash jain last updated on 27/May/20
	$characteric equation x^4 −3x^3 +2x^2 +2x−4=0 x^4 +x^3 −4x^3 −4x^2 +6x^2 +6x−4x−4=0 x^3 (x+1)−4x^2 (x+1)+6x(x+1)−4(x+1)=00 (x+1)(x^3 −4x^2 +6x−4)=0 (x+1)(x^3 −2x^2 −2x^2 +4x+2x−4)=0 (x+1){x^2 (x−2)−2x(x−2)+2(x−2)}=0 (x+1)(x−2)(x^2 −2x+2)=0 root,−1,2,1+i,1−i u_n =c_1 (−1)^n +c_2 2^n +c_3 (1+i)^n +c_4 (1−i)^n u_1 =3=−c_1 +2c_2 +c_3 (1+i)+c_4 (1−i) 3=−c_1 +2c_2 +c_3 (1+i)+c_4 (1−i) (I) u_2 =5=c_1 +4c_2 +c_3 (1+i)^2 +c_4 (1−i)^2 5=c_1 +4c_2 +2ic_3 −2ic_4 (II) u_3 =3=−c_1 +8c_2 +c_3 (1+i)^3 +c_4 (1−i)^3 3=−c_1 +8c_2 +2i(1+i)c_3 −2i(1−i)c_4 (III) u_4 =9=c_1 +16c_2 +c_3 (1+i)^4 +c_4 (1−i)^4 9=c_1 +16c_2 −4c_3 −4c_4 (IV) FROM (I), (II) ,(III) ,(IV) c_1 =1,c_2 =1,c_3 =1,c_4 =1 u_n =(−1)^n +2^n +(1+i)^n +(1−i)^n$
	$characteric equation$ $x^{4} - 3 x^{3} + 2 x^{2} + 2 x - 4 = 0$ $x^{4} + x^{3} - 4 x^{3} - 4 x^{2} + 6 x^{2} + 6 x - 4 x - 4 = 0$ $x^{3} (x + 1) - 4 x^{2} (x + 1) + 6 x (x + 1) - 4 (x + 1) = 00$ $(x + 1) (x^{3} - 4 x^{2} + 6 x - 4) = 0$ $(x + 1) (x^{3} - 2 x^{2} - 2 x^{2} + 4 x + 2 x - 4) = 0$ $(x + 1) {x^{2} (x - 2) - 2 x (x - 2) + 2 (x - 2)} = 0$ $(x + 1) (x - 2) (x^{2} - 2 x + 2) = 0$ $root, - 1, 2, 1 + i, 1 - i$ $u_{n} = c_{1} {(- 1)}^{n} + c_{2} 2^{n} + c_{3} {(1 + i)}^{n} + c_{4} {(1 - i)}^{n}$ $u_{1} = 3 = - c_{1} + 2 c_{2} + c_{3} (1 + i) + c_{4} (1 - i)$ $3 = - c_{1} + 2 c_{2} + c_{3} (1 + i) + c_{4} (1 - i) (I)$ $u_{2} = 5 = c_{1} + 4 c_{2} + c_{3} {(1 + i)}^{2} + c_{4} {(1 - i)}^{2}$ $5 = c_{1} + 4 c_{2} + 2 {i c}_{3} - 2 {i c}_{4} (I I)$ $u_{3} = 3 = - c_{1} + 8 c_{2} + c_{3} {(1 + i)}^{3} + c_{4} {(1 - i)}^{3}$ $3 = - c_{1} + 8 c_{2} + 2 i (1 + i) c_{3} - 2 i (1 - i) c_{4} (I I I)$ $u_{4} = 9 = c_{1} + 16 c_{2} + c_{3} {(1 + i)}^{4} + c_{4} {(1 - i)}^{4}$ $9 = c_{1} + 16 c_{2} - 4 c_{3} - 4 c_{4} (I V)$ $F R O M (I), (I I), (I I I), (I V)$ $c_{1} = 1, c_{2} = 1, c_{3} = 1, c_{4} = 1$ $u_{n} = {(- 1)}^{n} + 2^{n} + {(1 + i)}^{n} + {(1 - i)}^{n}$
	Commented by bobhans last updated on 27/May/20

	$great$
	Commented by PengagumRahasiamu last updated on 21/Jul/20
	Thank you Sir
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