Statement-1 : The variance of first $$n$$ even natural numbers is $$\frac{n^2 - 1}{4}$$. Statement-2 : The sum of first $$n$$ natural numbers is $$\frac{n(n+1)}{2}$$ and the sum of squares of first $$n$$ natural numbers is $$\frac{n(n+1)(2n+1)}{6}$$.

The first $$n$$ even natural numbers are $$2,\,4,\,6,\ldots,2n$$.

Mean of the data
Sum of the numbers $$\sum_{k=1}^{n} 2k = 2\sum_{k=1}^{n} k = 2 \cdot \frac{n(n+1)}{2} = n(n+1)$$
Mean $$\bar{x}= \frac{n(n+1)}{n}=n+1$$

Second moment about the origin
$$\sum_{k=1}^{n} (2k)^2 = 4\sum_{k=1}^{n} k^2 = 4 \cdot \frac{n(n+1)(2n+1)}{6}= \frac{2n(n+1)(2n+1)}{3}$$
Hence $$\overline{x^2}=\frac{1}{n}\sum (2k)^2= \frac{2(n+1)(2n+1)}{3}$$

Variance
Formula: $$\sigma^2 = \overline{x^2}-\bar{x}^2$$
$$\sigma^2 = \frac{2(n+1)(2n+1)}{3} - (n+1)^2$$ $$= (n+1)\left[\frac{2(2n+1)}{3} - (n+1)\right]$$ $$= (n+1)\left[\frac{4n+2 - 3n -3}{3}\right]$$ $$= (n+1)\left[\frac{n-1}{3}\right]$$ $$= \frac{n^2-1}{3}$$

Thus the correct variance is $$\frac{n^2-1}{3}$$, not $$\frac{n^2-1}{4}$$.
Statement-1 is therefore false.

Statement-2 merely lists the standard formulas $$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}, \qquad \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$$ which are correct. So Statement-2 is true.

Hence, Statement-1 is false while Statement-2 is true.
Option D which is: Statement-1 is false, Statement-2 is true