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Article #444;]
From: Karolina
Date: Sat, 23 Apr 2005 19:47
Date: Sat, 23 Apr 2005 19:47
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921 bytes
Ciag liczb "trojkatnych" jest zdefiniowany rekurencyjnie t1=1 tn+1=tn+(n+1) Wyznacz szesc poczatkowyc wyrazow tego ciagu
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