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a(n+1)=(n+1)²-5
Mam wykazac, ze:
a(n+1)-an>0 dla kazdego n naturalnego
a(n+1)-an=(n+1)²-5 -(n²-5)=n²+2n+1-5-n²+5==2n+1
a(n+1)-an=2n+1 >0 dla n∈N , a zatem jest to ciag rosnacy