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If have well-known formula $(n + 1) n / 2 = 1 + 2 +\cdots+ n$. If the difference between the closest numbers smaller, will obtain, for example $(n + 0,1) n / (2 \cdot 0,1) = 0,1 + 0,2 +\cdots+ n$. Now if the difference between the closest numbers the smallest possible, will obtain $(n + 0,0\ldots1) n / (2 \cdot 0,0\ldots 1) = 0,0\ldots1 + 0,0\ldots2 + \ldots + n$, so can conclude $n ^ 2 / 2 = (0,0\ldots1 + 0,0\ldots2 + \cdots + n) / 0,0\ldots1$ whether conclude is correct?


EDITED VERSION:

If have well-known formula $\frac{(n + 1)n}2 = 1 + 2 +\dots+ n$.

If the difference between the closest numbers smaller, will obtain, for example $\frac{(n + 0,1) n}{2.0,1} = 0,1 + 0,2 +\dots + n$.

Now if the difference between the closest numbers the smallest possible, will obtain $\frac{(n + 0,0\dots1) n}{2 . 0,0\dots 1} = 0,0\dots 1 + 0,0\dots 2 + \dots + n$ , so can conclude $\frac{n ^ 2}2 = \frac{0,0\dots1 + 0,0\dots2 + \dots + n}{0,0\dots1}$ whether conclude is correct?


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    possible duplicate of [Formula for the sum of the squares of numbers](http://math.stackexchange.com/questions/92662/formula-for-the-sum-of-the-squares-of-numbers)2012-01-10

1 Answers 1

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If we denote $k=n/\alpha$ then we get $\sum\limits_{i=1}^k i\alpha=\frac{k(k+1)}2 \alpha = \frac{k\alpha(k\alpha+\alpha)}{2\alpha} = \frac{n(n+\alpha)}{2\alpha}.$ For $\alpha$ of the form $10^{-s}$, i.e. $0,00\dots01$, this is precisely the first part of your question.

I am not sure I understand what you mean by the last part of your question, but if you want to say that $\sum\limits_{i=1}^k i\alpha=\frac{k(k+1)}2$ is approximately $\frac{n^2}{2\alpha}$, in some sense it is true, since the error is: $\frac{n(n+\alpha)}{2\alpha} - \frac{n^2}{2\alpha} = \frac {n\alpha}{2\alpha}=\frac n2,$ so the error has smaller order than the sum. (The error is $o(n^2)$, if you're familiar with this notation.)


If you know something about Riemann integral, you may notice that the sum $\alpha \sum\limits_{i=1}^k i\alpha$ is in fact upper Riemann sum for the function $f(x)=x$ on the interval $[0,n]$ (and for uniform partition of this interval). To see this just notice that the lengths of the intervals of partitions is $\alpha$ and the value at the end of the interval $[(i-1)\alpha,i\alpha]$ is $i\alpha$.

So it is expected that this sum is approximately $\frac{n^2}2$ for small values of $\alpha$. Your sum is $\alpha$-times smaller.