udowodnic, ze dla dowolnej liczby naturalnej zachodzi tozsamosc:
1+3+5+...+(2n-1)=n^2
1+3+5+ ...+(2n-1) = n²
a1 = 1, a2 = 3, a3 = 5, ... - ciag arytmetyczny
r = a2-a1 = 3-1 = 2
r = 2
n = n
Sn = [2a1+(n-1)r]·n/2
Sn = (2+2n-2)·n/2 = n·n = n²
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1+3+5+ ...+(2n-1) = n²
a1 = 1, a2 = 3, a3 = 5, ... - ciag arytmetyczny
r = a2-a1 = 3-1 = 2
r = 2
n = n
Sn = [2a1+(n-1)r]·n/2
Sn = (2+2n-2)·n/2 = n·n = n²