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Question Number 104544 by john santu last updated on 22/Jul/20

Given a_(n+1)  = 5a_n −6a_(n−1)   .If a_1 = 10 and a_2 = 26, find a_n ?

$$\mathcal{G}{iven}\:{a}_{{n}+\mathrm{1}} \:=\:\mathrm{5}{a}_{{n}} −\mathrm{6}{a}_{{n}−\mathrm{1}} \\ $$$$.\mathcal{I}{f}\:{a}_{\mathrm{1}} =\:\mathrm{10}\:{and}\:{a}_{\mathrm{2}} =\:\mathrm{26},\:{find}\:{a}_{{n}} ? \\ $$

Answered by bobhans last updated on 22/Jul/20

let a_n = Aρ^n ⇒Aρ^(n+1) −5Aρ^n +6Aρ^(n−1)  = 0  Aρ^(n−1) {ρ^2 −5ρ+6 } = 0  ⇒ρ= 3; 2 ⇒a_n = A.3^n +B.2^n   n=1 →3A +2B = 10  n =2 →9A +4B = 26  { ((A=2)),((B=2)) :}  a_n = 2.3^n  + 2.2^n  = 2.{3^n  + 2^n  } ■

$${let}\:{a}_{{n}} =\:{A}\rho^{{n}} \Rightarrow{A}\rho^{{n}+\mathrm{1}} −\mathrm{5}{A}\rho^{{n}} +\mathrm{6}{A}\rho^{{n}−\mathrm{1}} \:=\:\mathrm{0} \\ $$$${A}\rho^{{n}−\mathrm{1}} \left\{\rho^{\mathrm{2}} −\mathrm{5}\rho+\mathrm{6}\:\right\}\:=\:\mathrm{0} \\ $$$$\Rightarrow\rho=\:\mathrm{3};\:\mathrm{2}\:\Rightarrow{a}_{{n}} =\:{A}.\mathrm{3}^{{n}} +{B}.\mathrm{2}^{{n}} \\ $$$${n}=\mathrm{1}\:\rightarrow\mathrm{3}{A}\:+\mathrm{2}{B}\:=\:\mathrm{10} \\ $$$${n}\:=\mathrm{2}\:\rightarrow\mathrm{9}{A}\:+\mathrm{4}{B}\:=\:\mathrm{26}\:\begin{cases}{{A}=\mathrm{2}}\\{{B}=\mathrm{2}}\end{cases} \\ $$$${a}_{{n}} =\:\mathrm{2}.\mathrm{3}^{{n}} \:+\:\mathrm{2}.\mathrm{2}^{{n}} \:=\:\mathrm{2}.\left\{\mathrm{3}^{{n}} \:+\:\mathrm{2}^{{n}} \:\right\}\:\blacksquare \\ $$

Answered by OlafThorendsen last updated on 22/Jul/20

r^2 −5r+6 = 0  (r−3)(r−2) = 0  r_1  = 2 and r_2  = 3  a_n  = λ2^n +μ3^n   To find λ and μ   we use the initial conditions.  a_1  = 2λ+3μ = 10  a_2  = 4λ+9μ = 26  ⇒ λ = 2 and μ = 2  a_n  = 2.2^n +2.3^n  = 2^(n+1) +2.3^n

$${r}^{\mathrm{2}} −\mathrm{5}{r}+\mathrm{6}\:=\:\mathrm{0} \\ $$$$\left({r}−\mathrm{3}\right)\left({r}−\mathrm{2}\right)\:=\:\mathrm{0} \\ $$$${r}_{\mathrm{1}} \:=\:\mathrm{2}\:\mathrm{and}\:{r}_{\mathrm{2}} \:=\:\mathrm{3} \\ $$$${a}_{{n}} \:=\:\lambda\mathrm{2}^{{n}} +\mu\mathrm{3}^{{n}} \\ $$$$\mathrm{To}\:\mathrm{find}\:\lambda\:\mathrm{and}\:\mu\: \\ $$$$\mathrm{we}\:\mathrm{use}\:\mathrm{the}\:\mathrm{initial}\:\mathrm{conditions}. \\ $$$${a}_{\mathrm{1}} \:=\:\mathrm{2}\lambda+\mathrm{3}\mu\:=\:\mathrm{10} \\ $$$${a}_{\mathrm{2}} \:=\:\mathrm{4}\lambda+\mathrm{9}\mu\:=\:\mathrm{26} \\ $$$$\Rightarrow\:\lambda\:=\:\mathrm{2}\:\mathrm{and}\:\mu\:=\:\mathrm{2} \\ $$$${a}_{{n}} \:=\:\mathrm{2}.\mathrm{2}^{{n}} +\mathrm{2}.\mathrm{3}^{{n}} \:=\:\mathrm{2}^{{n}+\mathrm{1}} +\mathrm{2}.\mathrm{3}^{{n}} \\ $$

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