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    Question

    What is the average of the squares of the first 41

    natural numbers?
    A 567 Correct Answer Incorrect Answer
    B 574 Correct Answer Incorrect Answer
    C 581 Correct Answer Incorrect Answer
    D 588 Correct Answer Incorrect Answer
    E None of these Correct Answer Incorrect Answer

    Solution

    Sum of squares of first 'n' natural numbers = n ├Ч (n + 1) ├Ч (2n + 1) ├╖ 6
    So, average = n ├Ч (n + 1) ├Ч (2n + 1) ├╖ 6n = (n + 1) ├Ч (2n + 1) ├╖ 6
    = (41 + 1) ├Ч (2 ├Ч 41 + 1) ├╖ 6
    = 42 ├Ч 83 ├╖ 6
    = 581

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