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1 ^(2) + 2^(2) + 3^(2) + . . . + n^(2) =...

`1 ^(2) + 2^(2) + 3^(2) + . . . + n^(2) = (n (n + 1) (2 n + 1))/( 6)`

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The correct Answer is:
`= (n(n + 1) (2 n + 1))/( 6)` for all `n in N`
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