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?
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