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Re: [obm-l] soma da Eureka romena
- To: obm-l@xxxxxxxxxxxxxx
- Subject: Re: [obm-l] soma da Eureka romena
- From: Carlos Yuzo Shine <cyshine@xxxxxxxxx>
- Date: Mon, 19 Mar 2007 08:37:02 -0700 (PDT)
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hm...
Para facilitar a notação, seja w = \pi/n. Note que w não depende de k.
tg(w) = tg((k+1)w - kw) = [tg((k+1)w) - tg(kw)]/[1 + tg(kw)tg((k+1)w)]
Logo
1 + tg(kw)tg((k+1)w) = [tg((k+1)w) - tg(kw)]/tg(w),
ou seja,
tg(kw)tg((k+1)w) = [tg((k+1)w) - tg(kw)]/tg(w) - 1
e a soma S fica simples:
S = soma([tg((k+1)w) - tg(kw)]/tg(w) - 1)
= [tg(2w) - tg(w)]/tg(w) - 1 + [tg(3w) - tg(2w)]/tg(w) - 1 + ... + [tg(nw) - tg((n-1)w)]/tg(w) - 1
= [1/tg(w)][tg(2w) - tg(w) + tg(3w) - tg(2w) + ... + tg(nw) - tg((n-1)w)] - (n-1)
= [1/tg(w)][tg(nw) - tg(w)] - (n-1)
Mas tg(nw) = tg(\pi) = 0. Logo
S =[1/tg(w)][-tg(w)] - (n-1) = -1 - (n-1) = -n.
[]'s
Shine
----- Original Message ----
From: Luís Lopes <qed_texte@hotmail.com>
To: obm-l@mat.puc-rio.br
Sent: Monday, March 19, 2007 11:20:33 AM
Subject: [obm-l] soma da Eureka romena
Sauda,c~oes,
Esta é da Gazeta Matematica V.97, p.229.
Calcular
\sum_{k=1}^{n-1} \tan(k\pi/n) \tan[(k+1)\pi/n]
n>=3, ímpar.
[]'s
Luis
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