The sum of the series `1^2+3^2+5^2+.......\ n` terms is a.`(n(n+1)(2n+1))/2` b. `(n(2n-1)(2n+1))/3` c. `((n-1)^2(2n+1))/6` d. `((2n+1)^3)/3`

Explanation

To find the sum of the series \(1^2 + 3^2 + 5^2 + \ldots\) for \(n\) terms, we can follow these steps:

### Step 1: Identify the general term
The series consists of the squares of the first \(n\) odd numbers. The \(r\)-th odd number can be expressed as:
\[ a_r = 2r - 1 \]
Thus, the general term for the series is:
...