Sum of the series $1^{2} + 3^{2} + 5^{2} +$ ....... upto n terms is

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Solution

The correct option is A
$H e r e, a_{n} = [1 + (n - 1) 2]^{2} = (2 n - 1)^{2} = 4 n^{2} - 4 n + 1 ∴ S_{n} = 4 \sum n^{2} - 4 \sum n + n = 4 \frac{n (n + 1) (2 n + 1)}{6} - 4 \frac{n (n + 1)}{2} + n = \frac{n}{3} (4 n^{2} - 1)$

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\frac{3}{1^{2}} + \frac{5}{1^{2} + 2^{2}} + \frac{7}{1^{2} + 2^{2} + 3^{2}} + . . . .

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