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: 
Since you are already convinced that $\sum\_{i=1}^{n}a=na$ this might help:

$$\sum\_{i=0}^{n-1}a={a}+\sum\_{i=1}^{n-1}a=a+\sum\_{i=1}^{n}a-{a}=\sum\_{i=1}^{n}a.$$
You are summing $n$ terms, all equal to $a$. So
$$\sum\_{i=0}^{n-1} a = a+ a+\dotsb + a = na.$$