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Question Number 29148 by abdo imad last updated on 04/Feb/18

let give the sequence (u_n ) /u_0 =1 and u_1 =2 and ∀ n ∈N 2u_(n+2) =3 u_(n+1) −u_n . let give the sequence (v_n ) / v_n = u_(n+1) −u_n . 1) prove that (v_n ) is geometric .find v_n in terms of n 2) find u_n in terms of n.

$${let}\:{give}\:{the}\:{sequence}\:\left({u}_{{n}} \right)\:/{u}_{\mathrm{0}} =\mathrm{1}\:{and}\:{u}_{\mathrm{1}} =\mathrm{2}\:{and} \\ $$$$\forall\:{n}\:\in{N}\:\:\:\mathrm{2}{u}_{{n}+\mathrm{2}} =\mathrm{3}\:{u}_{{n}+\mathrm{1}} −{u}_{{n}} .\:{let}\:{give}\:{the}\:{sequence}\:\left({v}_{{n}} \right)\:/ \\ $$$${v}_{{n}} =\:{u}_{{n}+\mathrm{1}} −{u}_{{n}} \:. \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:\left({v}_{{n}} \right)\:{is}\:{geometric}\:.{find}\:{v}_{{n}} {in}\:{terms}\:{of}\:{n} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{u}_{{n}} \:{in}\:{terms}\:{of}\:{n}. \\ $$

Commented by abdo imad last updated on 08/Feb/18

1)v_(n+1) =u_(n+2) −u_(n+1) =(3/2)u_(n+1) −(1/2)u_n −u_(n+1) =(1/2)(u_(n+1) −u_n ) =(1/2) v_n so (v_n ) is a geometric progression with raizon q=(1/2) ⇒v_n =v_0 q^n but v_0 =u_1 −u_0 =1 ⇒ v_n =((1/2))^n 2) Σ_(k=0) ^(n−1) v_k = Σ_(k=0) ^(n−1) (u_(k+1) −u_k )=u_1 −u_0 +u_2 −u_1 +...+u_n −u_(n−1) =u_n −u_0 ⇒u_n =1+Σ_(k=0) ^(n−1) ((1/2))^k =1+((1−((1/2))^n )/(1−(1/2))) = 1+2(1−(1/2^n ))= 3 −(1/2^(n−1) ) u_n = 3−(1/2^(n−1) ) .

$$\left.\mathrm{1}\right){v}_{{n}+\mathrm{1}} ={u}_{{n}+\mathrm{2}} \:−{u}_{{n}+\mathrm{1}} =\frac{\mathrm{3}}{\mathrm{2}}{u}_{{n}+\mathrm{1}} \:−\frac{\mathrm{1}}{\mathrm{2}}{u}_{{n}} \:−{u}_{{n}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({u}_{{n}+\mathrm{1}} −{u}_{{n}} \right)\:=\frac{\mathrm{1}}{\mathrm{2}}\:{v}_{{n}} \:{so}\:\left({v}_{{n}} \right)\:{is}\:{a}\:{geometric}\:{progression} \\ $$$${with}\:{raizon}\:{q}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{v}_{{n}} ={v}_{\mathrm{0}} {q}^{{n}} \:\:{but}\:{v}_{\mathrm{0}} ={u}_{\mathrm{1}} −{u}_{\mathrm{0}} =\mathrm{1} \\ $$$$\Rightarrow\:{v}_{{n}} =\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} \\ $$$$\left.\mathrm{2}\right)\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} {v}_{{k}} =\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({u}_{{k}+\mathrm{1}} −{u}_{{k}} \right)={u}_{\mathrm{1}} −{u}_{\mathrm{0}} \:+{u}_{\mathrm{2}} −{u}_{\mathrm{1}} +...+{u}_{{n}} −{u}_{{n}−\mathrm{1}} \\ $$$$={u}_{{n}} \:−{u}_{\mathrm{0}} \:\Rightarrow{u}_{{n}} =\mathrm{1}+\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{k}} =\mathrm{1}+\frac{\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} }{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\:\mathrm{1}+\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\right)=\:\mathrm{3}\:−\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} } \\ $$$${u}_{{n}} =\:\mathrm{3}−\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }\:. \\ $$

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