$(2^{2} + 4^{2} + 6^{2} + . . . . . . . + 20^{2}) =$ ?

770

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1155

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1540

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385 \times 385

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Solution

The correct option is B $1540$
$(2^{2} + 4^{2} + 6^{2} + . . . . . . . + 20^{2}) = {(1 \times 2)}^{2} + {(2 \times 2)}^{2} + {(2 \times 3)}^{2} + . . . . . . + {(2 \times 10)}^{2}$
$= (2^{2} \times 1^{2}) + (2^{2} \times 2^{2}) + (2^{2} \times 3^{2}) + . . . . + (2^{2} \times 10^{2})$
$= (4 \times 5 \times 77)$
$= 1540$

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Similar questions

(1^{2} + 2^{2} + 3^{2} + . . . + 10^{2}) = 385

(2^{2} + 4^{2} + 6^{2} + . . . + 20^{2})

1^{2} + 2^{2} + 3^{2} + . . . . . . . {.10}^{2} = 385,

(2^{2} + 4^{4} + 6^{2} + . . . . . . . . {.20}^{2})

1^{2} + 2^{2} + 2^{2} = 3^{2}

2^{2} + 3^{2} + 6^{2} = 7^{2}

3^{2} + 4^{2} + 12^{2} = 13^{2}

4^{2} + 5^{2} + 20^{2} = 21^{2}

5^{2} + 6^{2} + 30^{2} = 31^{2}

6^{2} + 7^{2} + 42^{2} = . . .^{2}

2^{2} + 4^{2} + 6^{2} + . . . + 20^{2}

1^{2} + 3^{2} + 5^{2} + . . . + 19^{2}

\frac{1}{2^{2} - 1} + \frac{1}{(4^{2} - 1)} + \frac{1}{6^{2} - 1} + . . . . . + \frac{1}{20^{2} - 1}

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