Statement 1: The variance of first $$n$$ odd natural numbers is $$\frac{n^2-1}{3}$$. Statement 2: The sum of first $$n$$ odd natural number is $$n^2$$ and the sum of square of first $$n$$ odd natural numbers is $$\frac{n(4n^2+1)}{3}$$.

The sequence of the first $$n$$ odd natural numbers is $$1,\,3,\,5,\dots ,\,(2n-1)$$.

Step 1 - Mean of the data
Sum of the first $$n$$ odd numbers is known to be $$n^{2}$$, hence the mean is
$$\overline{x}= \frac{n^{2}}{n}=n$$.

Step 2 - Sum of the squares
Compute $$\displaystyle\sum_{k=1}^{n}(2k-1)^{2}$$.

$$\sum_{k=1}^{n}(2k-1)^{2}= \sum_{k=1}^{n}\bigl(4k^{2}-4k+1\bigr)$$
$$=4\sum_{k=1}^{n}k^{2}-4\sum_{k=1}^{n}k+\sum_{k=1}^{n}1$$

Using the standard formulae $$\sum k = \frac{n(n+1)}{2}$$ and $$\sum k^{2} = \frac{n(n+1)(2n+1)}{6}$$, we get

$$4\sum k^{2}=4\cdot\frac{n(n+1)(2n+1)}{6}= \frac{2n(n+1)(2n+1)}{3}$$
$$-4\sum k = -4\cdot\frac{n(n+1)}{2} = -2n(n+1)$$
$$\sum 1 = n$$

Adding them,

$$\sum_{k=1}^{n}(2k-1)^{2}= \frac{2n(n+1)(2n+1)}{3}-2n(n+1)+n$$
$$=n\left[\frac{4n^{2}+6n+2}{3}-2n-1\right]$$
$$=n\left[\frac{4n^{2}+6n+2-6n-3}{3}\right]$$
$$=n\left[\frac{4n^{2}-1}{3}\right]=\frac{n\,(4n^{2}-1)}{3}$$.

Therefore Statement 2’s second claim should be $$\frac{n(4n^{2}-1)}{3}$$, not $$\frac{n(4n^{2}+1)}{3}$$, so Statement 2 is false.

Step 3 - Variance
For a data set $$x_{1},x_{2},\dots ,x_{n}$$, the variance is
$$\sigma^{2}= \frac{1}{n}\sum_{i=1}^{n}x_{i}^{2}-\bigl(\overline{x}\bigr)^{2}$$.

Insert the values obtained:
$$\sigma^{2}= \frac{1}{n}\cdot\frac{n(4n^{2}-1)}{3}-n^{2}= \frac{4n^{2}-1}{3}-n^{2}$$
$$=\frac{4n^{2}-1-3n^{2}}{3}= \frac{n^{2}-1}{3}$$.

This equals the expression given in Statement 1, so Statement 1 is true.

Conclusion
Statement 1 is true, Statement 2 is false. Hence the correct choice is:

Option A which is: Statement 1 is true, Statement 2 is false.