Question Number 105205 by Ar Brandon last updated on 26/Jul/20

Commented byAziztisffola last updated on 26/Jul/20

ce n′est pas tres lisible.

Commented byAr Brandon last updated on 26/Jul/20

D'accord, désolé pour l'inconvenance.

Answered by mathmax by abdo last updated on 27/Jul/20

u_n =(3/4)u_(n−1)  +(1/4)u_(n−2)  ⇒4u_n =3u_(n−1)  +u_(n−2)  ⇒4u_(n+2) =3u_(n+1)  +u_n ⇒  4u_(n+2) −3u_(n+1) −u_n =0 →4r^2 −3r −1 =0  Δ =9−4(−4) =25 ⇒r_1 =((3+5)/8)=1 and r_2 =((3−5)/8) =−(1/4) ⇒  u_n =α +β(−(1/4))^n   u_0 =α+β  u_1 =α−(β/4) ⇒(5/4)β =u_0 −u_1   ⇒β =(4/5)(u_0 −u_1 )  α=u_0 −β =u_0 −(4/5)(u_o −u_1 ) =(1/5)u_0  +(4/5)u_1   ⇒u_n =(1/5)(u_0  +4u_1 )+(4/5)(u_0 −u_1 )(−(1/4))^n   if u_0 =u_1  ⇒u_n =u_0  ∀n so u_n  is constante  2)v_n =u_n −u_(n−1)  ⇒v_(n+1) =u_(n+1) −u_n =(3/4)u_n  +(1/4)u_(n−1) −u_(n−1)   =(3/4)u_n −(3/4)u_(n−1) =(3/4)(u_n −u_(n−1) ) =(3/4)v_n  ⇒v_n  is geometric with  q =(3/4) ⇒ v_n =v_1 ×((3/4))^(n−1)   v_1 =u_1 −u_0  ≠0 ⇒v_n =(u_1 −u_0 )((3/4))^(n−1)   Σ_(k=1) ^n  v_k =Σ_(k=1) ^n (u_k −u_(k−1) ) =u_1 −u_0  +u_2 −u_1  +...+u_n −u_(n−1)   =u_n −u_o  ⇒u_n −u_0 =(u_1 −u_0 )Σ_(k=1) ^n  ((3/4))^(k−1)      (k−1 =p)  =(u_1 −u_0 )Σ_(k=0) ^(n−1) ((3/4))^p   =(u_1 −u_0 )×((1−((3/4))^n )/(1−(3/4)))  =4(u_1 −u_0 )×(1−((3/4))^n ) ⇒u_n =u_0  +4(u_1 −u_0 )(1−((3/4))^n )

Commented byAr Brandon last updated on 27/Jul/20

Thanks Sir. Always willing to help ! 😃

Commented byabdomathmax last updated on 27/Jul/20

you are welcome sir.