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


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

$if a_{1} = 3, a_{2} = 2$ $a_{n + 2} = a_{n + 1} + \frac{a_{1}}{2}$ $find a_{6} = ?$ $mister W method$ $a_{n} = A {(\frac{1 + \sqrt{3}}{2})}^{n} + B {(\frac{1 - \sqrt{3}}{2})}^{n}$ $a_{1} = A (\frac{1 + \sqrt{3}}{2}) + B (\frac{1 - \sqrt{3}}{2}) = 3$ $a_{2} = A {(\frac{1 + \sqrt{3}}{2})}^{2} + B {(\frac{1 - \sqrt{3}}{2})}^{2} = 2$ $\Rightarrow A + B + (A - B) \sqrt{3} = 6$ $\Rightarrow 2 (A + B) + (A - B) \sqrt{3} = 4$ $A = \frac{4 - \sqrt{3}}{\sqrt{3}}, B = \frac{- 4 - \sqrt{3}}{3}$ $a_{n} = (\frac{4 - \sqrt{3}}{\sqrt{3}}) {(\frac{1 + \sqrt{3}}{2})}^{n} - (\frac{4 + \sqrt{3}}{\sqrt{3}}) {(\frac{1 - \sqrt{3}}{2})}^{n}$
Commented by jagoll last updated on 13/Feb/20

$mister W, my work is correct ?$
Commented by john santu last updated on 13/Feb/20

$yes$
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