The sum of $1^{2} + 3^{2} + 5^{2} . . . {.31}^{2}$ is

9455

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5456

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3468

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4892

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Solution

The correct option is B $5456$
$s = 1^{2} + 3^{2} + 5^{2} + - - - - - - 31^{2}$
$s = 1^{2} + 2^{2} + 3^{2} + 4^{2} + - - - 30^{2} + 31^{2} - (2^{2} + 4^{2} + - - - + 30^{2})$
we know $\sum^{2} x^{2} = {\frac{n (n - 1) (2 n + 1)}{6}}$
x=1
$s = 1^{2 + 2^{2} + 3^{2} + - - - - 31^{2 - 2^{2} (1^{2} + 2^{2} + 3^{2} + - - - + 15^{2})}}$
$s = \frac{31 (31 + 1) (62 + 1)}{6}$ $- \frac{4}{6} (15) (15 + 1) (30 + 1)$
$s = \frac{7}{6} [31 (32) (63) - 4 (15) (16) (31)]$
s=[5456.]

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