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Question
The sum of squares of first n natural numbers is given by `1/6n(n + 1)(2n + 1)` or `1/6(2n^3 + 3n^2 + n)`. Find the sum of squares of the first 10 natural numbers.
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Solution
Given, the sum of squares of first n natural numbers = `1/6(n +1)(2n + 1)`
∴ The sum of squares of first 10 natural numbers ...[∴ Put n = 10]
= `1/6 (10)(10 + 1)(2 xx 10+1)`
= `1/6 xx 10 xx 11 xx 21`
= 385
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