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Re: [obm-l] Potências
- To: obm-l@xxxxxxxxxxxxxx
- Subject: Re: [obm-l] Potências
- From: "Marcelo Salhab Brogliato" <msbrogli@xxxxxxxxx>
- Date: Tue, 10 Apr 2007 20:27:51 -0300
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- Reply-To: obm-l@xxxxxxxxxxxxxx
- Sender: owner-obm-l@xxxxxxxxxxxxxx
Ola,
veja:
(a + 1/a)^2 = a^2 + 1/a^2 + 2
assim: Sum (k=1 -> 27) (x^k + 1/x^k)^2 = Sum (k=1 ->27) (x^(2k) +
1/x^(2k) + 2) =
= Sum (k=1 ->27) (x^(2k) + 1/x^(2k)) + 54 =
= Sum (k=1 ->27) [x^(2k)] + Sum(k=1->27) [1/x^(2k)] + 54
opa.. temos 2 somatorios de PG finita, logo:
= x^2*[x^54 - 1]/[x^2 - 1] + (1/x^2)*[1/x^54 - 1]/[1/x^2 - 1] + 54
mas: (1/x^2)*[1/x^54 - 1]/[1/x^2 - 1] = [1/x^54 - 1]/[1-x^2] =
(1/x^54)*[1-x^54]/[1-x^2] =
= (1/x^54)*[x^54 - 1]/[x^2 - 1]
substituindo temos:
= x^2*[x^54 - 1]/[x^2 - 1] + (1/x^2)*[1/x^54 - 1]/[1/x^2 - 1] + 54 =
= x^2*(x^54 - 1)/(x^2 - 1) + (1/x^54)*(x^54 - 1)/(x^2 - 1) + 54 =
= (x^2 + 1/x^54)*(x^54 - 1)/(x^2 - 1) + 54
cara, nao sei c errei alguma conta.. mas acho q esta perto do fim já..
nao consegui fechar a questao.. dps eu tento de novo e se eu tiver alguma
ideia eu mando..
abracos,
Salhab
On 4/10/07, Fabio Honorato dos Santos <fabinho_g3@hotmail.com> wrote:
> [ OCM - 1997 ] Se x^2 + x + 1 = 0 , calcule o valor numérico de
>
> ( x + 1 / x )^2 + ( x^2 + 1 / x ^2)^2 + ... + ( x^27 + 1 / x ^27)^2 .
>
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