If $(1^{2} + 2^{2} + 3^{3} + . . . . {.12}^{2}) = 650$ , then the value of $(2^{2} + 4^{2} + 6^{2} + . . . . . . . + 24^{2})$ is

1300

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2600

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2500

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42250

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Solution

The correct option is B $2600$
Solve:

Given, $1^{2} + 2^{2} + 3^{2} + \dots + 12^{2} = 650$

$\Rightarrow (2^{2} + 4^{2} + 6^{2} + \dots {.24}^{2})$ we have to

find $2^{2} = 1^{2} \cdot 2^{2}$

$4^{2} = 2^{2} \cdot 2^{2}$

$6^{2} = 3^{2} \cdot 2^{2}$

$\Rightarrow (2^{2} + 4^{2} + 6^{2} + \dots .2 4^{2}) = 2^{2} (1^{2} + 2^{2} + 3^{2} + \dots 24^{2})$

$\Rightarrow (2^{2} + 4^{2} + 6^{2} + \dots 24^{2}) = 2^{2} \times 650$

$\Rightarrow 2^{2} + 4^{2} + 6^{2} + \dots 24^{2} = 650 \times 4$

$\Rightarrow 2^{2} + 4^{2} + 6^{2} + \dots 24^{2} = 2600$

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

=

\frac{1^{3} + 2^{3} + 3^{3} + 4^{3} + . . . . . . . {.12}^{3}}{1^{2} + 2^{2} + 3^{2} + 4^{2} + . . . . . {.12}^{2}} =

\frac{1^{2} + 2^{2}}{| 3 - -} + \frac{1^{2} + 2^{2} + 3^{2}}{| 4 - -} + \frac{1^{2} + 2^{2} + 3^{3} + 4^{2}}{| 5 - -} . . . . . . . .

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