# Property of summation

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.

• Sorry, I cannot figure out what is difficult to understand here. Can you enlighten me ? – Yves Daoust Mar 15 '19 at 14:29
• Recognize that $\{0,1,2,3,4,\dots,n-1\}$ has $n$ elements in it. If this is not immediately obvious why, then recognize that it has the $n-1$ positive elements $\{1,2,3,\dots,n-1\}$ and also the one additional zero element $\{0\}$. It follows that your summation is iterated a total of $n$ times (the one time when the index is zero, and then the following $n-1$ times while the index is positive for a total of $1+(n-1)=n$ times). – JMoravitz Mar 15 '19 at 14:29
• Thanks JMoravitz, your rationale was what I needed to convince myself. – Kolmogorovwannabe Mar 15 '19 at 14:40

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

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

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

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 $$n$$.

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

Hint:

$$a$$ times the number of terms.