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Sequence

Given that ...

Question

Given that $1^{2} + 2^{2} + 3^{2} + . . . . . . . {.10}^{2} = 385,$
then the value of $(2^{2} + 4^{4} + 6^{2} + . . . . . . . . {.20}^{2})$ is equal to

770

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1540

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1155

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{(385)}^{2}

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Solution

The correct option is B 1540
Let the sum of second series = m
Square of each number in second series is 4 times of square of each number of 1st series
Hence the Sum of square of numbers second series is 4times the sum of square of number of 1st series
As given Sum of 1st series = 385
$385 = \frac{m}{4}$
$m = 385 \times 4$
$\Rightarrow 1540$

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Similar questions

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

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

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

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

1^{2} + 2^{2} + 3^{2} + . . . . . . . . . + 512^{2} = m, t h e n

2^{2} + 4^{2} + 6^{2} + . . . . . . . . . + 1024^{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} +

= 21^{2}

5^{2} +

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

6^{2} + 7^{2} +

=

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