Question

the sum of squares of first n natural numbers is given by
$\frac{1}{6} n(n + 1)(2n + 1) \: or \: \frac{1}{6} ( {2n}^{3} + {3n}^{2} + n).$
find the sum of squares of the first ten natural numbers.
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Answer 1

Answer:

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Answer 2

Here,

$\bf S_n\to \tt\frac{1}{6} ( {2n}^{3} + {3n}^{2} + n) \\ \\ \bf S_n \to \tt{ \frac{ {2n}^{3} + {3n}^{2} + n }{6} }$

Now,

$\bf\to S_{10} = \tt \frac{ {2(10)}^{3} + {3(10)}^{2} + (10) }{6} \\ \\\bf\to S_{10} \tt = \frac{ {2(1000)}^{} + 3(100) + 10}{6} \\ \\\bf\to S_{10} \tt = \frac{2000 + 300 + 10}{6} \\ \\ \bf\to S_{10} \tt = \frac{2310}{6} \: \: \: \rm{or} \: \: \: \sf \red{385}$

Thus,

• Sum of squares of the first ten natural numbers is 385.