The value of $1^{2} + 3^{2} + 5^{2} + . . . . {.25}^{2}$ is:

1728

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1456

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2925

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1469

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Solution

The correct option is C $2925$
$1^{2} + 3^{2} + 5^{2} + . . . . + 25^{2}$
$\Rightarrow$ $(1^{2} + 2^{2} + 3^{2} + 4^{2} + . . . . + 25^{2}) - (2^{2} + 4^{2} + 6^{2} + 8^{2} . . . {.24}^{2})$

$\Rightarrow$ $(1^{2} + 2^{2} + 3^{2} + 4^{2} + . . {.25}^{2}) - [(2 \times 1)^{2} + (2 \times 2)^{2} + (2 \times 3)^{2} + . . . . (2 \times 12)^{2}]$

$\Rightarrow$ $(1^{2} + 2^{2} + 3^{2} + . . . . + 25^{2}) - 2^{2} (1^{2} + 2^{2} + 3^{2} + . . . . {.12}^{2})$

$1^{2} + 2^{2} + 3^{2} + . . . + n^{2} = \frac{n (n + 1) (2 n + 1)}{6}$

$\Rightarrow$ $\frac{25 (25 + 1) (2 \times 25 + 1)}{6} - 2^{2} \frac{12 (12 + 1) (2 \times 12 + 1)}{6}$

$\Rightarrow$ $\frac{25 \times 26 \times 51}{6} - 4 \times \frac{12 \times 13 \times 25}{6}$

$\Rightarrow$ $25 \times 13 \times 17 - 4 \times 2 \times 13 \times 25$

$\Rightarrow$ $25 \times 13 (17 - 8)$

$\Rightarrow$ $25 \times 13 \times 9$

$\Rightarrow$ $2925$

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