Question
What is the average of the squares of the first 41
natural numbers?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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