The sum of $10^{2} + 11^{2} + 12^{2} + . . . . + 20^{2}$ is

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Solution

Formula:
$1^{2} + 2^{2} + 3^{2} + \dots \dots + n^{2} = \frac{n (n - 1) (n - 2)}{6}$

$11^{2} + 12^{2} + \dots \dots + 20^{2}$

$= (1^{2} + 2^{2} + 3^{2} + \dots \dots + 20^{2}) - (1^{2} + 2^{2} + 3^{2} + \dots \dots + 10^{2})$

$= \frac{20 \times 21 \times 41}{6} - \frac{10 \times 11 \times 21}{6}$

$= 2870 - 385$

$= 2485$

$∴ 10^{2} + 11^{2} + \dots . + 20^{2} = 10^{2} + 2485 = 2585$

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