Find the sum of $1^{2} - 2^{2} + 3^{2} - 4^{2} + . . . . . . . . + 19^{2} - 20^{2} = . . . . . . .$

- 210

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120

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210

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- 120

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Solution

The correct option is B $- 210$
$1^{2} - 2^{2} + 3^{2} - 4^{2} + . . . . . . + 19^{2} - 20^{2} = (1^{2} + 3^{2} + . . . . . + 19^{2}) - (2^{2} + 4^{2} + . . . {.20}^{2}) = (1^{2} + 2^{2} + 3^{2} + 4^{2} + . . . . . . + 19^{2} + 20^{2}) - 2 (2^{2} + 4^{2} + . . . {.20}^{2}) = (\frac{20 \times 21 \times 41}{6}) - 8 \times (\frac{10 \times 11 \times 21}{6}) = 2870 - 3080 = - 210$

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Similar questions

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

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

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

(2^{2} + 4^{2} + 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}

∠ B O A = 210^{\circ}

∠ B O C = 120^{\circ}

∠ C O A .

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