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Prove:
1^2+2^2+3^2+...+n^2= [n(n+1)(2n+1)]/6
wher: n is natural number
Bosnian
answered
In google paste:
1^2+2^2+3^2+...+n^2= [n(n+1)(2n+1)]/6
when you see list of results go on:
h t t p s://socratic.o r g › questions › can-you-prove-that-1...
You will see solution with explanation.