Property of summation

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 1

In both cases – $$\sum_{i=0}^{n-1}a\quad \text { and }\quad\sum_{i=1}^{n}a$$

– there are exactly $n$ summands.

Answer 2

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 $na$.

Answer 3

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

Answer 4

Hint:

$a$ times the number of terms.

Answer 5

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

$$\sum_{i=0}^{n-1}a={a}+\sum_{i=1}^{n-1}a=a+\sum_{i=1}^{n}a-{a}=\sum_{i=1}^{n}a.$$

Answer 6

Since $a$ is not $i$-depending one can write: $$\sum_{i=0}^{n-1}{a}=a\sum_{i=0}^{n-1}{1}$$ And $\sum_{i=0}^{n-1}{1}=1+1+\cdots+1$ $n$ times which obviously is $n$.