Statement - I : The variance of first n even natural numbers is `(n^(2) - 1)/(4)`
Statement - II : The sum of first n natural numbers is `(n(n+1))/(2)` and the sum of the squares of first n natural numbers is `(n(n+1)(2n+1))/(6)`

Options

A. Statement-I is true, Statement-II is true, Statement -II is not a correct explanation for statement - I
B. Statement-I is true, Statement-II is false
C. Statement-I is false, Statement-II is true
D. Statement-I is true, Statement-II is true, Statement -II is a correct explanation for statement - I

Answer

To solve the problem, we need to analyze both statements provided in the question:

### Step 1: Analyze Statement I
**Statement I:** The variance of the first n even natural numbers is \((n^2 - 1)/4\).

1. **Identify the first n even natural numbers:**
The first n even natural numbers are \(2, 4, 6, \ldots, 2n\).

2. **Calculate the mean (\(\bar{x}\)):**
\[ \bar{x} = \frac{2 + 4 + 6 + \ldots + 2n}{n} = \frac{2(1 + 2 + 3 + \ldots + n)}{n} = \frac{2 \cdot \frac{n(n + 1)}{2}}{n} = n + 1 \]

3. **Calculate the variance (\(\sigma^2\)):**
The formula for variance is:
\[ \sigma^2 = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2 \]
We can also use the formula:
\[ \sigma^2 = \frac{1}{n} \left( \sum_{i=1}^{n} x_i^2 \right) - \bar{x}^2 \]

4. **Calculate \(\sum_{i=1}^{n} x_i^2\):**
The squares of the first n even numbers are:
\[ 2^2, 4^2, 6^2, \ldots, (2n)^2 = 4(1^2 + 2^2 + 3^2 + \ldots + n^2) = 4 \cdot \frac{n(n + 1)(2n + 1)}{6} \]
Therefore,
\[ \sum_{i=1}^{n} x_i^2 = \frac{2n(n + 1)(2n + 1)}{3} \]

5. **Substituting into the variance formula:**
\[ \sigma^2 = \frac{1}{n} \cdot \frac{2n(n + 1)(2n + 1)}{3} - (n + 1)^2 \]
Simplifying this gives:
\[ \sigma^2 = \frac{2(n + 1)(2n + 1)}{3} - (n^2 + 2n + 1) \]
\[ = \frac{2(n + 1)(2n + 1) - 3(n^2 + 2n + 1)}{3} \]
After further simplification, we find:
\[ = \frac{n^2 - 1}{3} \]

6. **Conclusion for Statement I:**
Since we derived that the variance is \((n^2 - 1)/3\) and not \((n^2 - 1)/4\), **Statement I is false.**

### Step 2: Analyze Statement II
**Statement II:** The sum of the first n natural numbers is \((n(n + 1))/2\) and the sum of the squares of the first n natural numbers is \((n(n + 1)(2n + 1))/6\).

1. **Sum of the first n natural numbers:**
\[ S_n = 1 + 2 + 3 + \ldots + n = \frac{n(n + 1)}{2} \]
This is a well-known formula and is **true.**

2. **Sum of the squares of the first n natural numbers:**
\[ S_{n^2} = 1^2 + 2^2 + 3^2 + \ldots + n^2 = \frac{n(n + 1)(2n + 1)}{6} \]
This is also a well-known formula and is **true.**

### Conclusion
- **Statement I is false.**
- **Statement II is true.**

Thus, the correct option is that **Statement I is false and Statement II is true.**

### Final Answer:
**Option C:** Statement I is false and Statement II is true.