a_1 =1 a_2 =−1 and a_n =−a_(n−1) −2a_(n−2) Find a


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Question Number 164956 by Jamshidbek last updated on 24/Jan/22

$a_{1} = 1 a_{2} = - 1 and a_{n} = - a_{n - 1} - {2 a}_{n - 2}$ $Find a_{n}$
	Answered by mr W last updated on 24/Jan/22
	$1,−1,0,1,−1,0,1,−1,0,... through “observation”: ⇒a_(3n) =0 ⇒a_(3n+1) =1 ⇒a_(3n+2) =−1 general method: r^2 +r+2=0 r=((−1±i(√3))/2)=−e^(±((πi)/3)) a_n =(−1)^n {Ae^((nπi)/3) +Be^(−((nπi)/3)) } a_2 =−a_1 −2a_0 ⇒−1=−1−2a_0 ⇒a_0 =0 a_0 =A+B=0 a_1 =−{A((1+i(√3))/2)+B((1−i(√3))/2)}=1 A{((1+i(√3))/2)−((1−i(√3))/2)}=−1 A=(i/( (√3)))=(1/( (√3)))e^((πi)/2) =−B a_n =(((−1)^n )/( (√3))){e^(((nπi)/3)+((πi)/2)) −e^(−((nπi)/3)+((πi)/2)) } a_n =(((−1)^n )/( (√3))){cos (((nπ)/3)+(π/2))−cos (−((nπ)/3)+(π/2))+i [sin (((nπ)/3)+(π/2))−sin (−((nπ)/3)+(π/2))]} a_n =(((−1)^n )/( (√3))){−2sin (((nπ)/3))+i [cos (((nπ)/3))−cos (((nπ)/3))} a_n =(((−1)^(n+1) 2 sin (((nπ)/3)))/( (√3)))$
	$1, - 1, 0, 1, - 1, 0, 1, - 1, 0, . . .$ $t h r o u g h ‘ ‘ {o b s e r v a t i o n}^{″} :$ $\Rightarrow a_{3 n} = 0$ $\Rightarrow a_{3 n + 1} = 1$ $\Rightarrow a_{3 n + 2} = - 1$ $g e n e r a l m e t h o d :$ $r^{2} + r + 2 = 0$ $r = \frac{- 1 \pm i \sqrt{3}}{2} = - e^{\pm \frac{π i}{3}}$ $a_{n} = {(- 1)}^{n} {{A e}^{\frac{n π i}{3}} + {B e}^{- \frac{n π i}{3}}}$ $a_{2} = - a_{1} - 2 a_{0} \Rightarrow - 1 = - 1 - 2 a_{0} \Rightarrow a_{0} = 0$ $a_{0} = A + B = 0$ $a_{1} = - {A \frac{1 + i \sqrt{3}}{2} + B \frac{1 - i \sqrt{3}}{2}} = 1$ $A {\frac{1 + i \sqrt{3}}{2} - \frac{1 - i \sqrt{3}}{2}} = - 1$ $A = \frac{i}{\sqrt{3}} = \frac{1}{\sqrt{3}} e^{\frac{π i}{2}} = - B$ $a_{n} = \frac{{(- 1)}^{n}}{\sqrt{3}} {e^{\frac{n π i}{3} + \frac{π i}{2}} - e^{- \frac{n π i}{3} + \frac{π i}{2}}}$ $a_{n} = \frac{{(- 1)}^{n}}{\sqrt{3}} {\cos (\frac{n π}{3} + \frac{π}{2}) - \cos (- \frac{n π}{3} + \frac{π}{2}) + i [\sin (\frac{n π}{3} + \frac{π}{2}) - \sin (- \frac{n π}{3} + \frac{π}{2})]}$ $a_{n} = \frac{{(- 1)}^{n}}{\sqrt{3}} {- 2 \sin (\frac{n π}{3}) + i [\cos (\frac{n π}{3}) - \cos (\frac{n π}{3})}$ $a_{n} = \frac{{(- 1)}^{n + 1} 2 \sin (\frac{n π}{3})}{\sqrt{3}}$
	Commented by solihin last updated on 24/Jan/22

	${h o w}^{'} d y o u g o t t h e a_{3 n} ?$
	Commented by solihin last updated on 24/Jan/22

	$i m e a n t h e s e q u e n c e$
	Commented by mr W last updated on 24/Jan/22

	$w i t h a_{1} = 1, a_{2} = - 1 a n d a_{n} = - a_{n - 1} - 2 a_{n - 2}$ $y o u g e t$ $a_{3} = 0$ $a_{4} = 1$ $a_{5} = - 1$ $a_{6} = 0$ $a_{7} = 1$ $a_{8} = - 1$ $a_{9} = 0$ $. . . .$ $t h e n y o u c a n s e e$ $a_{3 n} = 0$ $a_{3 n + 1} = 1$ $a_{3 n + 2} = - 1$ $t h i s i s n o t a n ‘ ‘ {e x a c t}^{″} m e t h o d . i t$ $i s b a s e d o n ‘ ‘ {o b s e r v a t i o n}^{″} .$ $t h e r e f o r e i h a v e g i v e n a s e c o n d a n d$ $m o r e g e n e r a l m e t h o d .$
	Commented by Tawa11 last updated on 24/Jan/22

	$Great sir .$
	Commented by Jamshidbek last updated on 25/Jan/22

	$Mistake$
	Commented by mr W last updated on 25/Jan/22

	$y o u s h o u l d b e m o r e c l e a r s i r! w h a t d o$ $y o u m e a n ? i s t h e r e a m i s t a k e i n t h e$ $q u e s t i o n o r i s t h e r e a m i s t a k e i n m y$ $s o l u t i o n o r w h a t e l s e ?$
	Commented by aleks041103 last updated on 25/Jan/22

	$I t h i n k t h a t y o u c a n a l s o p r o v e t h i s$ $b y i n d u c t i o n .$
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