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Question Number 90217 by Maclaurin Stickker last updated on 22/Apr/20

let x_(n )  be a sequence with x_0 =2 and x_1 =7  and x_(n+1) =7x_n −12x_(n−1) .  Find the general term of x_n .

$${let}\:{x}_{{n}\:} \:{be}\:{a}\:{sequence}\:{with}\:{x}_{\mathrm{0}} =\mathrm{2}\:{and}\:{x}_{\mathrm{1}} =\mathrm{7} \\ $$$${and}\:{x}_{{n}+\mathrm{1}} =\mathrm{7}{x}_{{n}} −\mathrm{12}{x}_{{n}−\mathrm{1}} . \\ $$$${Find}\:{the}\:{general}\:{term}\:{of}\:{x}_{{n}} . \\ $$

Commented by jagoll last updated on 22/Apr/20

p^2 −7p+12 = 0   p = 3 ∧ p = 4   x_n  = A.3^n  + B.4^n   n = 0 ⇒2 = A+B   n=1 ⇒ 7 = 3A+4B  B = 1 ∧ A = 1   ∴ x_n  = 1.3^n  + 1.4^n  = 3^n  + 4^n

$$\mathrm{p}^{\mathrm{2}} −\mathrm{7p}+\mathrm{12}\:=\:\mathrm{0}\: \\ $$$$\mathrm{p}\:=\:\mathrm{3}\:\wedge\:\mathrm{p}\:=\:\mathrm{4}\: \\ $$$$\mathrm{x}_{\mathrm{n}} \:=\:\mathrm{A}.\mathrm{3}^{\mathrm{n}} \:+\:\mathrm{B}.\mathrm{4}^{\mathrm{n}} \\ $$$$\mathrm{n}\:=\:\mathrm{0}\:\Rightarrow\mathrm{2}\:=\:\mathrm{A}+\mathrm{B}\: \\ $$$$\mathrm{n}=\mathrm{1}\:\Rightarrow\:\mathrm{7}\:=\:\mathrm{3A}+\mathrm{4B} \\ $$$$\mathrm{B}\:=\:\mathrm{1}\:\wedge\:\mathrm{A}\:=\:\mathrm{1}\: \\ $$$$\therefore\:\mathrm{x}_{\mathrm{n}} \:=\:\mathrm{1}.\mathrm{3}^{\mathrm{n}} \:+\:\mathrm{1}.\mathrm{4}^{\mathrm{n}} \:=\:\mathrm{3}^{\mathrm{n}} \:+\:\mathrm{4}^{\mathrm{n}} \\ $$

Commented by Maclaurin Stickker last updated on 22/Apr/20

thank you

$${thank}\:{you} \\ $$

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