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Question-31214

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Question Number 31214 by Tinkutara last updated on 03/Mar/18
Commented by abdo imad last updated on 03/Mar/18
we have u_(n+1) =3u_n −2 u_(n−1 ) ⇒u_(n+2) −3u_(n+1) +2u_n =0 the caracteristic equation is r^2 −3r +2=0 roots? Δ=9−8=1 ⇒r_1 =((3+1)/2)=2 and r_2 =((3−1)/2)=1 so the roots are simple ⇒u_n =ar_1 ^n +br_2 ^n =1+a 2^n we have u_0 =1+a=2 ⇒a=1 and we verify thst u_1 =3 ⇒ u_n =1+2^n .
$${we}\:{have}\:{u}_{{n}+\mathrm{1}} =\mathrm{3}{u}_{{n}} −\mathrm{2}\:{u}_{{n}−\mathrm{1}\:} \Rightarrow{u}_{{n}+\mathrm{2}} \:−\mathrm{3}{u}_{{n}+\mathrm{1}} \:+\mathrm{2}{u}_{{n}} =\mathrm{0} \\ $$$${the}\:{caracteristic}\:{equation}\:{is}\:{r}^{\mathrm{2}} \:−\mathrm{3}{r}\:+\mathrm{2}=\mathrm{0}\:{roots}? \\ $$$$\Delta=\mathrm{9}−\mathrm{8}=\mathrm{1}\:\Rightarrow{r}_{\mathrm{1}} =\frac{\mathrm{3}+\mathrm{1}}{\mathrm{2}}=\mathrm{2}\:{and}\:{r}_{\mathrm{2}} =\frac{\mathrm{3}−\mathrm{1}}{\mathrm{2}}=\mathrm{1}\:{so}\:{the}\:{roots} \\ $$$${are}\:{simple}\:\Rightarrow{u}_{{n}} ={ar}_{\mathrm{1}} ^{{n}} \:+{br}_{\mathrm{2}} ^{{n}} =\mathrm{1}+{a}\:\mathrm{2}^{{n}} \:{we}\:{have} \\ $$$${u}_{\mathrm{0}} =\mathrm{1}+{a}=\mathrm{2}\:\Rightarrow{a}=\mathrm{1}\:{and}\:{we}\:{verify}\:{thst}\:{u}_{\mathrm{1}} =\mathrm{3}\:\Rightarrow \\ $$$${u}_{{n}} =\mathrm{1}+\mathrm{2}^{{n}} . \\ $$
Commented by Tinkutara last updated on 04/Mar/18
Thanks Sir!
Answered by math solver last updated on 03/Mar/18
(2) 2^n +1.
$$\left(\mathrm{2}\right)\:\mathrm{2}^{\mathrm{n}} +\mathrm{1}. \\ $$
Answered by ajfour last updated on 03/Mar/18
u_(n+1) =3(3u_(n−1) −2u_(n−2) )−2u_(n−1) =7u_(n−1) −6u_(n−2) =7(3u_(n−2) −2u_(n−3) )−6u_(n−2) =15u_(n−2) −14u_(n−3) = 15(3u_(n−3) −2u_(n−4) )−14u_(n−3) = 31u_(n−3) −30u_(n−4) =(2^5 −1)u_(n−3) −(2^5 −2)u_(n−4) =(2^(n+1) −1)u_(n−(n−1)) −(2^(n+1) −2)u_(n−n) = (2^(n+1) −1)×3−(2^(n+1) −2)×2 = 2^(n+1) +1 ⇒ u_n = 2^n +1 .
$${u}_{{n}+\mathrm{1}} =\mathrm{3}\left(\mathrm{3}{u}_{{n}−\mathrm{1}} −\mathrm{2}{u}_{{n}−\mathrm{2}} \right)−\mathrm{2}{u}_{{n}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{7}{u}_{{n}−\mathrm{1}} −\mathrm{6}{u}_{{n}−\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{7}\left(\mathrm{3}{u}_{{n}−\mathrm{2}} −\mathrm{2}{u}_{{n}−\mathrm{3}} \right)−\mathrm{6}{u}_{{n}−\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{15}{u}_{{n}−\mathrm{2}} −\mathrm{14}{u}_{{n}−\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\mathrm{15}\left(\mathrm{3}{u}_{{n}−\mathrm{3}} −\mathrm{2}{u}_{{n}−\mathrm{4}} \right)−\mathrm{14}{u}_{{n}−\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\mathrm{31}{u}_{{n}−\mathrm{3}} −\mathrm{30}{u}_{{n}−\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:=\left(\mathrm{2}^{\mathrm{5}} −\mathrm{1}\right){u}_{{n}−\mathrm{3}} −\left(\mathrm{2}^{\mathrm{5}} −\mathrm{2}\right){u}_{{n}−\mathrm{4}} \\ $$$$\:\:\:=\left(\mathrm{2}^{{n}+\mathrm{1}} −\mathrm{1}\right){u}_{{n}−\left({n}−\mathrm{1}\right)} −\left(\mathrm{2}^{{n}+\mathrm{1}} −\mathrm{2}\right){u}_{{n}−{n}} \\ $$$$\:\:\:=\:\left(\mathrm{2}^{{n}+\mathrm{1}} −\mathrm{1}\right)×\mathrm{3}−\left(\mathrm{2}^{{n}+\mathrm{1}} −\mathrm{2}\right)×\mathrm{2} \\ $$$$\:\:=\:\mathrm{2}^{{n}+\mathrm{1}} +\mathrm{1} \\ $$$$\Rightarrow\:\:\:\:\:\boldsymbol{{u}}_{\boldsymbol{{n}}} =\:\mathrm{2}^{\boldsymbol{{n}}} +\mathrm{1}\:\:. \\ $$
Commented by Tinkutara last updated on 03/Mar/18
Thanks for it!

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