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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.

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    $\begingroup$ Sorry, I cannot figure out what is difficult to understand here. Can you enlighten me ? $\endgroup$ – Yves Daoust Mar 15 at 14:29
  • $\begingroup$ 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). $\endgroup$ – JMoravitz Mar 15 at 14:29
  • $\begingroup$ Thanks JMoravitz, your rationale was what I needed to convince myself. $\endgroup$ – Kolmogorovwannabe Mar 15 at 14:40
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Since you are allready convinced that $\sum_{i=1}^{n}a=na$ this might help:

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

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

- there are exactly $n$ summands.

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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 $n$.

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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 $na$.

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

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Hint:

$a$ times the number of terms.

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