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[obm-l] Somatório
- To: obm-l@xxxxxxxxxxxxxx
- Subject: [obm-l] Somatório
- From: "Felipe Régis" <bluttau@xxxxxxxxx>
- Date: Sat, 5 May 2007 22:36:25 -0300
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- Reply-To: obm-l@xxxxxxxxxxxxxx
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Olá pessoal,
Alguém poderia me ajudar a demonstrar que,
S(n) = Sum[i=1->n] {i/[(i+1)(i+2)(i+3)]} = [n(n+3)]/[4(n+1)(n+2)]
Comecei a desenvolver a soma isoladamente mas não achei nenhuma relação que pudesse me ajudar:
S(0)=0
S(1)=1/24
S(2)= 3/40
S(3)=1/10
...
S(n)= [n(n+3)]/[4(n+1)(n+2)]
Obrigado!
Felipe Régis e Silva