Proving an identity and some interesting maths stuff

[gentsagree]

So, I would like to prove that

$$\gamma^{\mu_{1}...\mu_{r}}=(-)^{r(r-1)/2}\gamma^{\mu_{r}...\mu_{1}}$$

where the matrix gamma is a totally antisymmetric matrix defined as $\gamma^{\mu_{1}...\mu_{r}}=\gamma^{[\mu_{1}}\gamma^{\mu_{2}}...\gamma^{\mu_{r}]}$

What I have done is to prove that

$$\gamma^{\mu_{1}...\mu_{r}}=(-)^{(r-1)+(r-2)+...+1}\gamma^{\mu_{r}...\mu_{1}}$$

by simply commuting all the matrices past each other until their order is reversed (picking up just the minus sign as they are antisymmetrised, so we can take $\mu_{i}\neq\mu_{j}$ for $i\neq j$).

What's a nice way to see that $(r-1)+(r-2)+...+1=r(r-1)/2$? It works for some values of r, which one can see by substituting in.

ALSO - PART 2

I am aware of $\sum_{n=1}^{\infty}n=\frac{x(x+1)}{2}=-\frac{1}{12}$,

but I found out that
$$\int^{1}_{0}\frac{x(x-1)}{2}dx=-\frac{1}{12}$$

Any comments or clarifications on this relationship between $\frac{x(x-1)}{2}$ and $\frac{x(x+1)}{2}$.

 

[Avodyne]

You're just summing the positive integers from 1 to $r-1$. Here's how Gauss solved this problem (for $r=101$) when he was in elementary school:
http://mathandmultimedia.com/2010/09/15/sum-first-n-positive-integers/