If $1^{2} + 2^{2} + 3^{2} + . . . . . . . . . + 512^{2} = m, t h e n$
$2^{2} + 4^{2} + 6^{2} + . . . . . . . . . + 1024^{2}$ is equal to

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m^{2}

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m^{3}

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Solution

The correct option is B 4m
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 = m
so Sum of second series = 4m

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Similar questions

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

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

\frac{1}{2^{2}} + \frac{1}{4^{2}} + \frac{1}{6^{2}} + . . . . . . \infty = \frac{π^{2}}{24}

\frac{1}{1^{2}} + \frac{1}{3^{2}} + \frac{1}{5^{2}} + . . . . . . \infty

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} + . . . . . . = . . . . . . . .

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} +

=

1^{2} - 2^{2} + 3^{2} - 4^{2} + 5^{2} - 6^{2} + . . . . . . . . .

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