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u-1-2-u-n-1-3u-n-2-u-n-n-




Question Number 199709 by tri26112004 last updated on 08/Nov/23
u_1  = 2   u_(n+1)  = 3u_n  + 2  u_n  → n ¿
$${u}_{\mathrm{1}} \:=\:\mathrm{2}\: \\ $$$${u}_{{n}+\mathrm{1}} \:=\:\mathrm{3}{u}_{{n}} \:+\:\mathrm{2} \\ $$$${u}_{{n}} \:\rightarrow\:{n}\:¿ \\ $$
Answered by mr W last updated on 08/Nov/23
u_(n+1) +1=3u_n +2+1=3(u_n +1)  u_n +1=3(u_(n−1) +1)=3^2 (u_(n−2) +1)=...=3^(n−1) (u_1 +1)=3^(n−1) (2+1)=3^n   ⇒u_n =3^n −1
$${u}_{{n}+\mathrm{1}} +\mathrm{1}=\mathrm{3}{u}_{{n}} +\mathrm{2}+\mathrm{1}=\mathrm{3}\left({u}_{{n}} +\mathrm{1}\right) \\ $$$${u}_{{n}} +\mathrm{1}=\mathrm{3}\left({u}_{{n}−\mathrm{1}} +\mathrm{1}\right)=\mathrm{3}^{\mathrm{2}} \left({u}_{{n}−\mathrm{2}} +\mathrm{1}\right)=…=\mathrm{3}^{{n}−\mathrm{1}} \left({u}_{\mathrm{1}} +\mathrm{1}\right)=\mathrm{3}^{{n}−\mathrm{1}} \left(\mathrm{2}+\mathrm{1}\right)=\mathrm{3}^{{n}} \\ $$$$\Rightarrow{u}_{{n}} =\mathrm{3}^{{n}} −\mathrm{1} \\ $$

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