algebra precalculus - Formula for the sum of the squares of numbers

We have the well-known formula

$$\frac{n (n + 1) (2 n + 1)}{6} = 1^2 + 2^2 + \cdots + n^2 .$$

If the difference between the closest numbers is smaller, we obtain, for example

$$\frac{n \times (n + 0.1) (2 n + 0.1) }{6 \cdot 0.1} = 0.1^2 + 0.2^2 + \cdots + n^2 .$$

It is easy to check. Now if the difference between the closest numbers becomes smallest possible, we will obtain

$$\frac{n \cdot (n + 0.0..1) \cdot (2 n + 0.0..1)}{6 \cdot 0.0..1} = 0.0..1^2 + 0.0..2^2 + \cdots + n^2$$

So can conclude that

$$\frac{2n ^ 3}{6} = \frac{n ^ 3}{3} = \frac{0.0..1 ^ 2 + 0.0..2 ^ 2 + \cdots + n ^ 2}{0.0..1}.$$

Is this conclusion correct?