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 × (n + 1) × (2n + 1) ÷ 6
    So, average = n × (n + 1) × (2n + 1) ÷ 6n = (n + 1) × (2n + 1) ÷ 6
    = (41 + 1) × (2 × 41 + 1) ÷ 6
    = 42 × 83 ÷ 6
    = 581

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