Question Number 149551 by puissant last updated on 06/Aug/21

Answered by Olaf_Thorendsen last updated on 06/Aug/21

a_n = P(n) = an^2 +bn+c ∀n∈N, a_(n+1) = 3a_n −n^2 +n ⇔ a(n+1)^2 +b(n+1)+c = 3an^2 +3bn+3c−n^2 +n ⇔ (2a−1)n^2 +(−2a+2b+1)n−a−b+2c = 0 ⇔ { ((2a−1 = 0)),((−2a+2b+1 = 0)),((−a−b+2c = 0)) :} ⇔ a = (1/2), b = 0, c = (1/4) a_n = P(n) = (1/2)n^2 +(1/4)

Commented bypuissant last updated on 06/Aug/21

merci monsieur Olaf...

Commented bypuissant last updated on 06/Aug/21

de^� sole^� monsieur mais c′est a_(n+1) .. car c′est floue vraiment de^� sole^� .. mais je vais rectifier.. de^� sole^� ...

Commented bypuissant last updated on 06/Aug/21

et moi je trouve a_n =(1/2)n^2 +(1/4)..

Commented byOlaf_Thorendsen last updated on 06/Aug/21

j′ai rectifie

Commented bypuissant last updated on 06/Aug/21

merci..