Question: Very short question. Could you please explain me why

$$\sum\_{i=0}^{n-1} a = na$$
with $a$ a constant? 
I know that 

$$\sum\_{i=1}^{n} a = na$$

but in my case the sum starts from zero and finishes for $(n-1)$.

Thanks.

Answer: 
What is the definition of $\sum\_{i=0}^{n-1} x\_i$? It is exactly $x\_0+x\_1+...x\_{n-1}$. If $x\_0=x\_1=...x\_{n-1}=a$ then it means you just sum $a$ $n$ times. And that gives $na$.
You are summing $n$ terms, all equal to $a$. So
$$\sum\_{i=0}^{n-1} a = a+ a+\dotsb + a = na.$$