1^2 + 2^2 + 3^2 + ... + n^2 = ____?
- n(n+1)/2
- n(n+1)(2n+1)/6
- n(n-1)/2
- None of these
Explanation
The sum of squares of first n natural numbers is given by:
1^2 + 2^2 + 3^2 + ... + n^2 = n(n+1)(2n+1)/6
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