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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)

$${a}_{{n}} \:=\:\mathrm{P}\left({n}\right)\:=\:{an}^{\mathrm{2}} +{bn}+{c} \\ $$$$\forall{n}\in\mathbb{N},\:{a}_{{n}+\mathrm{1}} \:=\:\mathrm{3}{a}_{{n}} −{n}^{\mathrm{2}} +{n} \\ $$$$\Leftrightarrow\:{a}\left({n}+\mathrm{1}\right)^{\mathrm{2}} +{b}\left({n}+\mathrm{1}\right)+{c}\:= \\ $$$$\mathrm{3}{an}^{\mathrm{2}} +\mathrm{3}{bn}+\mathrm{3}{c}−{n}^{\mathrm{2}} +{n} \\ $$$$\Leftrightarrow\:\left(\mathrm{2}{a}−\mathrm{1}\right){n}^{\mathrm{2}} +\left(−\mathrm{2}{a}+\mathrm{2}{b}+\mathrm{1}\right){n}−{a}−{b}+\mathrm{2}{c}\:=\:\mathrm{0} \\ $$$$\Leftrightarrow\:\begin{cases}{\mathrm{2}{a}−\mathrm{1}\:=\:\mathrm{0}}\\{−\mathrm{2}{a}+\mathrm{2}{b}+\mathrm{1}\:=\:\mathrm{0}}\\{−{a}−{b}+\mathrm{2}{c}\:=\:\mathrm{0}}\end{cases} \\ $$$$\Leftrightarrow\:{a}\:=\:\frac{\mathrm{1}}{\mathrm{2}},\:{b}\:=\:\mathrm{0},\:{c}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${a}_{{n}} \:=\:\mathrm{P}\left({n}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}{n}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}} \\ $$

Commented by puissant last updated on 06/Aug/21

merci monsieur Olaf...

$$\mathrm{merci}\:\mathrm{monsieur}\:\mathrm{Olaf}... \\ $$

Commented by puissant last updated on 06/Aug/21

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

$$\mathrm{et}\:\mathrm{moi}\:\mathrm{je}\:\mathrm{trouve}\:\mathrm{a}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{n}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}.. \\ $$

Commented by Olaf_Thorendsen last updated on 06/Aug/21

j′ai rectifie

$$\mathrm{j}'\mathrm{ai}\:\mathrm{rectifie} \\ $$

Commented by puissant last updated on 06/Aug/21

merci..

$$\mathrm{merci}.. \\ $$

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