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Question Number 217190 by efronzo1 last updated on 05/Mar/25

Given a_(n+1) = a_n + a_(n+2) where a_3 = 4 and a_5 = 6 find a_n .

$$\mathrm{Given}\:\mathrm{a}_{\mathrm{n}+\mathrm{1}} \:=\:\mathrm{a}_{\mathrm{n}} \:+\:\mathrm{a}_{\mathrm{n}+\mathrm{2}} \: \\ $$$$\:\:\mathrm{where}\:\mathrm{a}_{\mathrm{3}} =\:\mathrm{4}\:\mathrm{and}\:\mathrm{a}_{\mathrm{5}} =\:\mathrm{6} \\ $$$$\:\mathrm{find}\:\mathrm{a}_{\mathrm{n}} \:. \\ $$

Answered by mr W last updated on 05/Mar/25

a_(n+2) −a_(n+1) +a_n =0 characteristic equation: r^2 −r+1=0 ⇒r_(1,2) =((1±i(√3))/2)=e^(±((πi)/3)) say a_n =Ae^((nπi)/3) +Be^(−((nπi)/3)) or a_n =C cos ((nπ)/3)+D sin ((nπ)/3) a_3 =−C=4 ⇒C=−4 a_5 =−4 cos ((5π)/3)+D sin ((5π)/3)=6 ⇒D=−((16)/( (√3))) ⇒a_n =−4(cos ((nπ)/3)+(4/( (√3))) sin ((nπ)/3)) or a_n =−((4(√(57)))/( 3)) sin (((nπ)/3)+tan^(−1) ((√3)/4))

$${a}_{{n}+\mathrm{2}} −{a}_{{n}+\mathrm{1}} +{a}_{{n}} =\mathrm{0} \\ $$$${characteristic}\:{equation}: \\ $$$${r}^{\mathrm{2}} −{r}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{r}_{\mathrm{1},\mathrm{2}} =\frac{\mathrm{1}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}={e}^{\pm\frac{\pi{i}}{\mathrm{3}}} \\ $$$${say}\:{a}_{{n}} ={Ae}^{\frac{{n}\pi{i}}{\mathrm{3}}} +{Be}^{−\frac{{n}\pi{i}}{\mathrm{3}}} \\ $$$${or}\:{a}_{{n}} ={C}\:\mathrm{cos}\:\frac{{n}\pi}{\mathrm{3}}+{D}\:\mathrm{sin}\:\frac{{n}\pi}{\mathrm{3}} \\ $$$${a}_{\mathrm{3}} =−{C}=\mathrm{4}\:\Rightarrow{C}=−\mathrm{4} \\ $$$${a}_{\mathrm{5}} =−\mathrm{4}\:\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{3}}+{D}\:\mathrm{sin}\:\frac{\mathrm{5}\pi}{\mathrm{3}}=\mathrm{6}\:\Rightarrow{D}=−\frac{\mathrm{16}}{\:\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow{a}_{{n}} =−\mathrm{4}\left(\mathrm{cos}\:\frac{{n}\pi}{\mathrm{3}}+\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}\:\mathrm{sin}\:\frac{{n}\pi}{\mathrm{3}}\right) \\ $$$${or} \\ $$$${a}_{{n}} =−\frac{\mathrm{4}\sqrt{\mathrm{57}}}{\:\mathrm{3}}\:\mathrm{sin}\:\left(\frac{{n}\pi}{\mathrm{3}}+\mathrm{tan}^{−\mathrm{1}} \frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\right) \\ $$

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