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soit-u-1-4-u-o-1-n-N-montrer-par-reccurence-que-U-n-2-2U-n-1-U-n-




Question Number 166379 by SANOGO last updated on 19/Feb/22
soit{_(u_1 =4) ^(u_o =1)   ∀nεN  montrer par reccurence  que:  U_(n+2) =2U_(n+1) −U_n
$${soit}\left\{_{{u}_{\mathrm{1}} =\mathrm{4}} ^{{u}_{{o}} =\mathrm{1}} \:\:\forall{n}\epsilon{N}\right. \\ $$$${montrer}\:{par}\:{reccurence}\:\:{que}: \\ $$$${U}_{{n}+\mathrm{2}} =\mathrm{2}{U}_{{n}+\mathrm{1}} −{U}_{{n}} \\ $$
Answered by qaz last updated on 19/Feb/22
U_(n+2) =2U_(n+1) −U_n   ⇒λ^2 −2λ+1=0  ⇒λ_1 =λ_2 =1  ⇒U_n =C_1 +nC_2   U_0 =1 Λ U_1 =4  ⇒U_n =1+3n
$$\mathrm{U}_{\mathrm{n}+\mathrm{2}} =\mathrm{2U}_{\mathrm{n}+\mathrm{1}} −\mathrm{U}_{\mathrm{n}} \\ $$$$\Rightarrow\lambda^{\mathrm{2}} −\mathrm{2}\lambda+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\lambda_{\mathrm{1}} =\lambda_{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow\mathrm{U}_{\mathrm{n}} =\mathrm{C}_{\mathrm{1}} +\mathrm{nC}_{\mathrm{2}} \\ $$$$\mathrm{U}_{\mathrm{0}} =\mathrm{1}\:\Lambda\:\mathrm{U}_{\mathrm{1}} =\mathrm{4} \\ $$$$\Rightarrow\mathrm{U}_{\mathrm{n}} =\mathrm{1}+\mathrm{3n} \\ $$

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