Let `S_(n)` denote the sum of n terms of the series
`1^(2)+3xx2^(2)+3^(2)+3xx4^(2)+5^(2)+3xx6^(2)+7^(2)+ . . . .`
Statement -1: If n is odd, then `S_(n)=(n(n+1)(4n-1))/(6)`
Statement -2: If n is even, then `S_(n)=(n(n+1)(4n+5))/(6)`

Options

A. Statement -1 is true, Statement -2 is True, Statement -2 is a correct explanation for Statement for Statement -1.
B. Statement -1 is true, Statement -2 is True, Statement -2 is not a correct explanation for Statement for Statement -1.
C. Statement -1 is true, Statement -2 is False.
D. Statement -1 is False, Statement -2 is True.

Explanation

To solve the problem, we need to find the sum \( S_n \) of the first \( n \) terms of the series given by:

\[ 1^2 + 3 \times 2^2 + 3^2 + 3 \times 4^2 + 5^2 + 3 \times 6^2 + 7^2 + \ldots \]

### Step 1: Identify the Pattern in the Series

The series can be broken down into two types of terms:
1. Terms of the form \( k^2 \) where \( k \) is an odd integer.
2. Terms of the form \( 3 \times (2m)^2 \) where \( m \) is a positive integer.

### Step 2: Separate the Terms Based on Odd and Even \( n \)

#### Case 1: \( n \) is Even

Let \( n = 2m \). The sum \( S_n \) can be expressed as:

\[ S_n = 1^2 + 3^2 + 5^2 + \ldots + (2m-1)^2 + 3 \times (2^2 + 4^2 + 6^2 + \ldots + (2m)^2) \]

The first part is the sum of squares of the first \( m \) odd numbers, and the second part is three times the sum of squares of the first \( m \) even numbers.

Using the formulas:
- The sum of the squares of the first \( m \) odd numbers is \( \frac{m(2m-1)(2m+1)}{3} \).
- The sum of the squares of the first \( m \) even numbers is \( \frac{m(m+1)(2m+1)}{3} \).

Thus, we can simplify \( S_n \) for even \( n \):

\[ S_n = \frac{m(2m-1)(2m+1)}{3} + 3 \times \frac{m(m+1)(2m+1)}{3} \]

Combining these gives:

\[ S_n = \frac{m(2m-1)(2m+1) + 3m(m+1)(2m+1)}{3} \]

This simplifies to:

\[ S_n = \frac{n(n+1)(4n+5)}{6} \]

#### Case 2: \( n \) is Odd

Let \( n = 2m + 1 \). The sum \( S_n \) can be expressed as:

\[ S_n = 1^2 + 3^2 + 5^2 + \ldots + (2m-1)^2 + 3 \times (2^2 + 4^2 + 6^2 + \ldots + (2m)^2) + (2m+1)^2 \]

Following similar steps as before, we find:

\[ S_n = \frac{m(2m-1)(2m+1)}{3} + 3 \times \frac{m(m+1)(2m+1)}{3} + (2m+1)^2 \]

This simplifies to:

\[ S_n = \frac{n(n+1)(4n-1)}{6} \]

### Final Result

Thus, we have:

- If \( n \) is odd, then \( S_n = \frac{n(n+1)(4n-1)}{6} \).
- If \( n \) is even, then \( S_n = \frac{n(n+1)(4n+5)}{6} \).