If $(1^{2} + 2^{2} + 3^{2} + . . . + 10^{2}) = 385$ then the value of $(2^{2} + 4^{2} + 6^{2} + . . . + 20^{2})$

770

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1155

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1540

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(385 x 385)

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Solution

The correct option is C 1540
$(2^{2} + 4^{2} + 6^{2} + . . . + 20^{2}) = {(1 \times 2)}^{2} + {(2 \times 2)}^{2} + {(2 \times 3)}^{2} + . . . + {(2 \times 10)}^{2}$

$= (2^{2} \times 1^{2}) + (2^{2} \times 2^{2}) + (2^{2} \times 3^{2}) + . . . + (2^{2} \times 10^{2})$
$= 2^{2} \times [1^{2} + 2^{2} + 3^{2} + . . . + 10^{2}]$
$(1^{2} + 2^{2} + 3^{2} + . . . . + n^{2}) = \frac{1}{6} n (n + 1) (2 n + 1)$
$= (4 \times \frac{1}{6} \times 10 \times 11 \times 21)$
$= (4 \times 5 \times 77)$
$= 1540$

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