probability - Why does $sumlimits_{i=1}^n frac{i-1}{n} = frac{n-1}{2}$?

I am trying to understand the accepted answer to this question: Find: The expected number of urns that are empty

And am stuck on the part I mentioned above. I understand that:

$\sum\limits_{i=1}^n \frac{i-1}{n}=\frac{0}{n}+\frac{1}{n}+...+\frac{n-1}{n}.$ But why does this equal $\frac{n\choose2}{n}$?

I have also seen this sum solved as follows:

$\sum\limits_{i=1}^n \frac{i-1}{n}=\frac{1}{n}\sum\limits_{i=1}^n i-1=\frac{1}{n}\sum\limits_{j=0}^{n-1}j=\frac{1}{n}\frac{n(n-1)}{2}=\frac{n-1}{2}$

With this method, I am having trouble seeing what exactly $j$ is and why we all of sudden entered it into the equation.

I'd appreciate if someone could shed light on either of these ways of computing the sum.

Thanks

Answer

The equation

$$\sum_{i=1}^n(i-1)=\sum_{j=0}^{n-1}j\tag{1}$$

is just a change of variable. Let $j=i-1$; then $i=j+1$, so the lefthand side of $(1)$ becomes

$$\sum_{j+1=1}^nj\;,\tag{2}$$

When $j+1=1$, clearly $j=0$, and when $j+1=n$, $j=n-1$. Thus, as $j+1$ runs from $1$ up through $n$, $j$ itself runs from $0$ up through $n-1$, and we can rewrite $(2)$ as the righthand side of $(1)$.

An alternative approach that does not involve changing the index variable is to decompose the sum:

$$\sum_{i=1}^n(i-1)=\sum_{i=1}^ni-\sum_{i=1}^n1=\frac{n(n+1)}2-n=\frac{n^2+n-2n}2=\frac{n^2-n}2=\frac{n(n-1)}2\;,$$

where the first summation uses the familiar formula for the sum of the first $n$ positive integers, and the second is just the sum of $n$ ones, or $n$.