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The arithemetic mean of the squares of first n natural numbers is `"________"`.A. `((n+1) (2n+1))/(6)`B. `(n+1)/(6)`C. `(n^(2) - 1)/(6)`D. `(n-1)/(6)` |
Answer» Correct Answer - a `sum n^(2) = (n(n+1) (2n+1))/(6)` AM = `(n(n+1) (2n+1))/(6n) = ((n+1) (2n+1))/(6)` |
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