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∑i = 0+1+2+…+n = n(n+1)/2
∑i2 = 0+12+22+…+n2 = n(n+1)(2n+1)/6
∑i3 = 0+13+23+…+n3 = (n(n+1)/2)2 = (∑i)2
...more sums of series...
Binomial theorem
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