Question Number 164956 by Jamshidbek last updated on 24/Jan/22
$$\mathrm{a}_{\mathrm{1}} =\mathrm{1}\:\mathrm{a}_{\mathrm{2}} =−\mathrm{1}\:\:\mathrm{and}\:\:\mathrm{a}_{\mathrm{n}} =−\mathrm{a}_{\mathrm{n}−\mathrm{1}} −\mathrm{2a}_{\mathrm{n}−\mathrm{2}} \\ $$$$\mathrm{Find}\:\:\mathrm{a}_{\mathrm{n}} \\ $$
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)))](Q164963.png)
$$\mathrm{1},−\mathrm{1},\mathrm{0},\mathrm{1},−\mathrm{1},\mathrm{0},\mathrm{1},−\mathrm{1},\mathrm{0},... \\ $$$${through}\:``{observation}'': \\ $$$$\Rightarrow{a}_{\mathrm{3}{n}} =\mathrm{0} \\ $$$$\Rightarrow{a}_{\mathrm{3}{n}+\mathrm{1}} =\mathrm{1} \\ $$$$\Rightarrow{a}_{\mathrm{3}{n}+\mathrm{2}} =−\mathrm{1} \\ $$$$ \\ $$$${general}\:{method}: \\ $$$${r}^{\mathrm{2}} +{r}+\mathrm{2}=\mathrm{0} \\ $$$${r}=\frac{−\mathrm{1}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}=−{e}^{\pm\frac{\pi{i}}{\mathrm{3}}} \\ $$$${a}_{{n}} =\left(−\mathrm{1}\right)^{{n}} \left\{{Ae}^{\frac{{n}\pi{i}}{\mathrm{3}}} +{Be}^{−\frac{{n}\pi{i}}{\mathrm{3}}} \right\} \\ $$$${a}_{\mathrm{2}} =−{a}_{\mathrm{1}} −\mathrm{2}{a}_{\mathrm{0}} \:\Rightarrow−\mathrm{1}=−\mathrm{1}−\mathrm{2}{a}_{\mathrm{0}} \:\Rightarrow{a}_{\mathrm{0}} =\mathrm{0} \\ $$$${a}_{\mathrm{0}} ={A}+{B}=\mathrm{0} \\ $$$${a}_{\mathrm{1}} =−\left\{{A}\frac{\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}+{B}\frac{\mathrm{1}−{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right\}=\mathrm{1} \\ $$$${A}\left\{\frac{\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{1}−{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right\}=−\mathrm{1} \\ $$$${A}=\frac{{i}}{\:\sqrt{\mathrm{3}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}{e}^{\frac{\pi{i}}{\mathrm{2}}} =−{B} \\ $$$${a}_{{n}} =\frac{\left(−\mathrm{1}\right)^{{n}} }{\:\sqrt{\mathrm{3}}}\left\{{e}^{\frac{{n}\pi{i}}{\mathrm{3}}+\frac{\pi{i}}{\mathrm{2}}} −{e}^{−\frac{{n}\pi{i}}{\mathrm{3}}+\frac{\pi{i}}{\mathrm{2}}} \right\} \\ $$$${a}_{{n}} =\frac{\left(−\mathrm{1}\right)^{{n}} }{\:\sqrt{\mathrm{3}}}\left\{\mathrm{cos}\:\left(\frac{{n}\pi}{\mathrm{3}}+\frac{\pi}{\mathrm{2}}\right)−\mathrm{cos}\:\left(−\frac{{n}\pi}{\mathrm{3}}+\frac{\pi}{\mathrm{2}}\right)+{i}\:\left[\mathrm{sin}\:\left(\frac{{n}\pi}{\mathrm{3}}+\frac{\pi}{\mathrm{2}}\right)−\mathrm{sin}\:\left(−\frac{{n}\pi}{\mathrm{3}}+\frac{\pi}{\mathrm{2}}\right)\right]\right\} \\ $$$${a}_{{n}} =\frac{\left(−\mathrm{1}\right)^{{n}} }{\:\sqrt{\mathrm{3}}}\left\{−\mathrm{2sin}\:\left(\frac{{n}\pi}{\mathrm{3}}\right)+{i}\:\left[\mathrm{cos}\:\left(\frac{{n}\pi}{\mathrm{3}}\right)−\mathrm{cos}\:\left(\frac{{n}\pi}{\mathrm{3}}\right)\right\}\right. \\ $$$${a}_{{n}} =\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \mathrm{2}\:\mathrm{sin}\:\left(\frac{{n}\pi}{\mathrm{3}}\right)}{\:\sqrt{\mathrm{3}}} \\ $$
Commented by solihin last updated on 24/Jan/22
$$ \\ $$$$ \\ $$$${how}'{d}\:\:{you}\:\:{got}\:\:{the}\:\:{a}_{\mathrm{3}{n}} ? \\ $$$$ \\ $$
Commented by solihin last updated on 24/Jan/22
$${i}\:\:{mean}\:\:{the}\:\:{sequence} \\ $$
Commented by mr W last updated on 24/Jan/22
$${with}\:{a}_{\mathrm{1}} =\mathrm{1},\:{a}_{\mathrm{2}} =−\mathrm{1}\:{and}\:{a}_{{n}} =−{a}_{{n}−\mathrm{1}} −\mathrm{2}{a}_{{n}−\mathrm{2}} \\ $$$${you}\:{get} \\ $$$${a}_{\mathrm{3}} =\mathrm{0} \\ $$$${a}_{\mathrm{4}} =\mathrm{1} \\ $$$${a}_{\mathrm{5}} =−\mathrm{1} \\ $$$${a}_{\mathrm{6}} =\mathrm{0} \\ $$$${a}_{\mathrm{7}} =\mathrm{1} \\ $$$${a}_{\mathrm{8}} =−\mathrm{1} \\ $$$${a}_{\mathrm{9}} =\mathrm{0} \\ $$$$.... \\ $$$${then}\:{you}\:{can}\:{see} \\ $$$${a}_{\mathrm{3}{n}} =\mathrm{0} \\ $$$${a}_{\mathrm{3}{n}+\mathrm{1}} =\mathrm{1} \\ $$$${a}_{\mathrm{3}{n}+\mathrm{2}} =−\mathrm{1} \\ $$$${this}\:{is}\:{not}\:{an}\:``{exact}''\:{method}.\:{it} \\ $$$${is}\:{based}\:{on}\:``{observation}''. \\ $$$${therefore}\:{i}\:{have}\:{given}\:{a}\:{second}\:{and} \\ $$$${more}\:{general}\:{method}. \\ $$
Commented by Tawa11 last updated on 24/Jan/22
$$\mathrm{Great}\:\mathrm{sir}. \\ $$
Commented by Jamshidbek last updated on 25/Jan/22
$$\mathrm{Mistake} \\ $$
Commented by mr W last updated on 25/Jan/22
$${you}\:{should}\:{be}\:{more}\:{clear}\:{sir}!\:{what}\:{do} \\ $$$${you}\:{mean}?\:{is}\:{there}\:{a}\:{mistake}\:{in}\:{the} \\ $$$${question}\:{or}\:{is}\:{there}\:{a}\:{mistake}\:{in}\:{my} \\ $$$${solution}\:{or}\:{what}\:{else}? \\ $$
Commented by aleks041103 last updated on 25/Jan/22
$${I}\:{think}\:{that}\:{you}\:{can}\:{also}\:{prove}\:{this} \\ $$$${by}\:{induction}. \\ $$