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Re: [obm-l] função
- To: obm-l@xxxxxxxxxxxxxx
- Subject: Re: [obm-l] função
- From: "Igor Battazza" <battazza@xxxxxxxxx>
- Date: Thu, 27 Mar 2008 02:39:10 -0300
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2008/3/26, Marcus Aurelio <marcusaurelio80@xxxxxxxxx>:
>
>
>
>
> Alguém pode me ajudar nessa
>
> f(x) + f(x-1) = x^2 se f(19) = 94 calcule f(94)?
>
>
Olá
f(x) = x^2 - f(x - 1), então f(94) = 94^2 - f(93),
f(93) = 93^2 - f(92),
f(92) = 92^2 - f(91).
....
f(20) = 20^2 - f(19)
Perceba que substituindo os valores, teremos a seguinte soma: f(94) =
(94^2 - 93^2) + (92^2 - 91^2) + ... + (22^2 - 21^2) + 20^2 - 94
mas (n + 1)^2 - n^2 = [(n + 1) - n][(n + 1) + n], portanto a soma fica:
f(94) = 1*(94 + 93) + 1*(92 + 91) + 1*(90 + 89) + ... + 20^2 - 94 ->
-> f(94) = (94 + 93 + 92 + ... + 21) + 20^2 - 94 = 4225 + 400 - 94 = 4561
(a soma de 21 até 94 é uma P.A. de razão 1)
"acho" que é isso
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