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


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

$G i v e n a_{n + 1} = 5 a_{n} - 6 a_{n - 1}$ $. I f a_{1} = 10 a n d a_{2} = 26, f i n d a_{n} ?$
	Answered by bobhans last updated on 22/Jul/20
	$let a_n = Aρ^n ⇒Aρ^(n+1) −5Aρ^n +6Aρ^(n−1) = 0 Aρ^(n−1) {ρ^2 −5ρ+6 } = 0 ⇒ρ= 3; 2 ⇒a_n = A.3^n +B.2^n n=1 →3A +2B = 10 n =2 →9A +4B = 26 { ((A=2)),((B=2)) :} a_n = 2.3^n + 2.2^n = 2.{3^n + 2^n } ■$
	$l e t a_{n} = A ρ^{n} \Rightarrow A ρ^{n + 1} - 5 A ρ^{n} + 6 A ρ^{n - 1} = 0$ $A ρ^{n - 1} {ρ^{2} - 5 ρ + 6} = 0$ $\Rightarrow ρ = 3; 2 \Rightarrow a_{n} = A . 3^{n} + B . 2^{n}$ $n = 1 \to 3 A + 2 B = 10$ $n = 2 \to 9 A + 4 B = 26 {\begin{cases} A = 2 \\ B = 2 \end{cases}$ $a_{n} = 2 . 3^{n} + 2 . 2^{n} = 2 . {3^{n} + 2^{n}} ◼$
	Answered by OlafThorendsen last updated on 22/Jul/20

	$r^{2} - 5 r + 6 = 0$ $(r - 3) (r - 2) = 0$ $r_{1} = 2 and r_{2} = 3$ $a_{n} = λ 2^{n} + μ 3^{n}$ $To find λ and μ$ $we use the initial conditions .$ $a_{1} = 2 λ + 3 μ = 10$ $a_{2} = 4 λ + 9 μ = 26$ $\Rightarrow λ = 2 and μ = 2$ $a_{n} = 2 . 2^{n} + 2 . 3^{n} = 2^{n + 1} + 2 . 3^{n}$
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