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Article #445

Re: ;]

#445
From: Damian Sobota
Date: Sat, 23 Apr 2005 19:50
11 lines
1118 bytes
Karolina napisaล‚(a):
> Ciag liczb "trojkatnych" jest zdefiniowany rekurencyjnie
> t1=1
> tn+1=tn+(n+1)
> Wyznacz szesc poczatkowyc wyrazow tego ciagu

To jakis zart? Sadzac po temacie to chyba tak.

-- 
Pozdrawiam,
Damian Sobota.
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From: Damian Sobota <dam.usun.sob@wp.tezusun.pl>
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Subject: Re: ;]
Date: Sat, 23 Apr 2005 21:50:04 +0200
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