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Article #448

Re: ;]

#448
From: Maciek
Date: Mon, 25 Apr 2005 09:25
12 lines
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Użytkownik "Karolina" <karcia17@poczta.onet.pl> napisał
w wiadomości news:d4e8ku$hir$1@news.onet.pl...
> Ciag liczb "trojkatnych" jest zdefiniowany rekurencyjnie
> t1=1
> tn+1=tn+(n+1)
> Wyznacz szesc poczatkowyc wyrazow tego ciagu

Wszystkie one są wymienione w trójkącie Pascala.


Maciek 
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From: "Maciek" <maciek@elkomtech.com.pl.nospam>
Newsgroups: alt.pl.matematyka
Subject: Re: ;]
Date: Mon, 25 Apr 2005 11:25:17 +0200
Organization: Bractwo Smocze
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