the sum of the squares I think it is interesting, if we have the  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 (let's call is a) we obtain, for example, if a=0.1
$$\frac{n  (n + 0.1)  (2 n + 0.1) }{6 \cdot 0.1} = 0.1^2 + 0.2^2 + \cdots + n^2 .$$ 
or, as another example if  a = 0.01 .$$\frac{n  (n + 0.01)  (2 n + 0.01) }{6 \cdot 0.01} = 0.01^2 + 0.02^2 + \cdots + n^2 .$$  
Now if we follow the same logic, I suppose if the difference between the closest numbers becomes smallest possible (a = 0.0...1), we will obtain 
$$ \frac{n (n + 0.0..1) (2 n  + 0.0..1)}{6 \cdot 0.0..1} = 0.0..1^2 + 0.0..2^2 + \cdots + n^2$$ 
So can conclude that
$$\frac{2n ^ 3}{6} = \frac{n ^ 3}{3} = \ (0.0..1 ^ 2 + 0.0..2 ^ 2 .. + n ^ 2) * {0.0..1}$$ If we take that,  a = 0.0...1 and,  b = $\frac {n}{0.0...1}$ and we rearrange the formula we will get this:

$$\frac{n ^ 3}{3} = 1^2a^3 + 2^2a^3 + \dots+b^2 a^3$$

following the same logic we prove that the formula holds for each degree (it is easy to check), so that we could generalize, If we take that degree call m, we will get

$$\frac{n ^ m}{m} = 1^{m-1}a^m + 2^{m-1}a^m + \dots+b^{m-1} a^m$$

so the question is this expression can be interpreted as a Riemann sum, and why?
 A: Indeed, if $\delta=\frac 1m$ then
$$ \frac{n(n+\delta)(2n+\delta)}{6}=\frac1{m^3}\frac{mn(mn+1)(2mn+1)}{6}=\frac1{m^3}\sum_{k=1}^{mn}k^2=\frac1{m}\sum_{k=1}^{mn}\left(\frac km\right)^2.$$
However, there is no such thing as "smalles possible" and from that point on you get some factros wrong.
What you really get in the limit as $m\to\infty$ is that
$$ \frac{n^3}{3}=\int_0^nx^2\,\mathrm dx.$$
This is because the last expression can be interpreted as a Riemann sum.
A: The quantity on the right of your last equation can be expressed as 
$$ R(n, \epsilon) = 
\sum_{k=1}^{n/\epsilon} k^2 \epsilon^3 $$ where $\epsilon = 0.00\ldots 01$ approaches zero.
Yore conclusion is correct in the following sense:
$$\lim_{\epsilon \to 0^+} \frac{R(n, \epsilon)}{n^3/3} = 1 $$
A: I think this is correct, as you are essentially integrating $x^2$, to get the result of $\frac{x^3}3$. this, if you don't know, is adding together rectangles of increasingly smaller width (a) and heights which are the result of a function ($x^2$).
