0

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.

MarianD
  • 2,953
Kolmogorovwannabe
  • 311
  • 2
  • 11
  • 7
    Sorry, I cannot figure out what is difficult to understand here. Can you enlighten me ? –  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

6 Answers6

24

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

– there are exactly $n$ summands.

enter image description here

MarianD
  • 2,953
8

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

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

Ernie060
  • 6,073
drhab
  • 151,093
3

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 $n$.

MarianD
  • 2,953
DINEDINE
  • 6,081
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 $na$.

Mark
  • 39,605
1

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

Hugo
  • 1,609
0

Hint:

$a$ times the number of terms.