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Question Number 81458 by jagoll last updated on 13/Feb/20

if a_1 = 3 ,a_2 =2 a_(n+2) = a_(n+1) +(a_1 /2) find a_6 =? mister W method a_n =A(((1+(√3))/2))^n +B(((1−(√3))/2))^n a_1 = A(((1+(√3))/2))+B(((1−(√3))/2))=3 a_2 = A(((1+(√3))/2))^2 +B(((1−(√3))/2))^2 =2 ⇒A+B+(A−B)(√3) =6 ⇒2(A+B)+(A−B)(√3) =4 A= ((4−(√3))/(√3)) , B = ((−4−(√3))/3) a_n = (((4−(√3))/(√3)))(((1+(√3))/2))^n −(((4+(√3))/(√3)))(((1−(√3))/2))^n

$$\mathrm{if}\:\mathrm{a}_{\mathrm{1}} \:=\:\mathrm{3}\:,\mathrm{a}_{\mathrm{2}} =\mathrm{2} \\ $$$$\mathrm{a}_{\mathrm{n}+\mathrm{2}} \:=\:\mathrm{a}_{\mathrm{n}+\mathrm{1}} +\frac{\mathrm{a}_{\mathrm{1}} }{\mathrm{2}} \\ $$$$\mathrm{find}\:\mathrm{a}_{\mathrm{6}} \:=? \\ $$$$\mathrm{mister}\:\mathrm{W}\:\mathrm{method} \\ $$$$\mathrm{a}_{\mathrm{n}} \:=\mathrm{A}\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{n}} +\mathrm{B}\left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{n}} \\ $$$$\mathrm{a}_{\mathrm{1}} =\:\mathrm{A}\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)+\mathrm{B}\left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)=\mathrm{3} \\ $$$$\mathrm{a}_{\mathrm{2}} \:=\:\mathrm{A}\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{B}\left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{2} \\ $$$$\Rightarrow\mathrm{A}+\mathrm{B}+\left(\mathrm{A}−\mathrm{B}\right)\sqrt{\mathrm{3}}\:=\mathrm{6} \\ $$$$\Rightarrow\mathrm{2}\left(\mathrm{A}+\mathrm{B}\right)+\left(\mathrm{A}−\mathrm{B}\right)\sqrt{\mathrm{3}}\:=\mathrm{4} \\ $$$$\mathrm{A}=\:\frac{\mathrm{4}−\sqrt{\mathrm{3}}}{\sqrt{\mathrm{3}}}\:,\:\mathrm{B}\:=\:\frac{−\mathrm{4}−\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$$\mathrm{a}_{\mathrm{n}} \:=\:\left(\frac{\mathrm{4}−\sqrt{\mathrm{3}}}{\sqrt{\mathrm{3}}}\right)\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{n}} −\left(\frac{\mathrm{4}+\sqrt{\mathrm{3}}}{\sqrt{\mathrm{3}}}\right)\left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{n}} \\ $$

Commented by jagoll last updated on 13/Feb/20

mister W, my work is correct?

$$\mathrm{mister}\:\mathrm{W},\:\mathrm{my}\:\mathrm{work}\:\mathrm{is}\:\mathrm{correct}? \\ $$

Commented by john santu last updated on 13/Feb/20

yes

$$\mathrm{yes} \\ $$

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