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I've a fomula that looks like this:

How to properly represent this formula using SUM Symbol and MEAN symbol? $$ r=2 \left(3(f_1) +\frac{(g_1+...+ g_n)}{n} + \frac{s_1 +... + s_n}{n}\right) + x_1 + y_1 $$

I don't know how to represent this in math symbols. I know that the hole form is a sum and i also know that i'm calculating the mean whem i perform a sum of n numbers and i divide it by n.

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  • $\begingroup$ You divide the sum of two elements only, seemingly, twice. Or, did you mean $g_1+g_2+g_3+\dots+g_n$ instead of '(g1 + gN)'? $\endgroup$ – Berci Jan 5 '14 at 20:21
  • $\begingroup$ yeah you get it right: i mean g1+g2+g3+⋯+gn $\endgroup$ – Lothre1 Jan 5 '14 at 20:22
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    $\begingroup$ Is the N in gN supposed to be the same as the n in the denominator? So that (g1 + … + gN)/n is the mean of the g's? $\endgroup$ – MJD Jan 5 '14 at 20:25
  • $\begingroup$ yes. you were right $\endgroup$ – Lothre1 Jan 5 '14 at 20:45
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$$ \frac{(g_1+...+g_N)}{n}=\frac{\sum_{i=1}^{n} g_i}{n} $$ $$ \frac{(s_1+...+s_N)}{n}=\frac{\sum_{i=1}^{n} s_i}{n} $$ so,

$$ r=2\left(3(f_1)+2 \left( \frac{\sum_{i=1}^{n} g_i}{n}\right)+\left(\frac{\sum_{i=1}^{n} s_i}{n}\right)\right)+x_1+y_1 $$

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  • $\begingroup$ it seems to me that it have some mistakes. Doesn't it? ((3 * f1) + (2* ( (g1 + … + gN) / n) + ((s1 + … + sN) / n)) * 2 the hole formula is multiplied by 2 which is not represented above. (2* ( (g1 + … + gN) / n) -> this is multiplied by 2 ((s1 + … + sN) / n)) -> this is NOT multiplied by 2. That's just a mean. $\endgroup$ – Lothre1 Jan 5 '14 at 20:43
  • $\begingroup$ Yes. It's correct. Thank you very much Felipe $\endgroup$ – Lothre1 Jan 5 '14 at 20:54
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The mean of some elements $g_1,g_2,\dots,g_n$ is often denoted by $\bar g$, you can introduce it by a sum notation. $$\bar g:=\frac{\sum_{k=1}^ng_k}n=\frac{g_1+g_2+\dots+g_n}n\,.$$

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  • $\begingroup$ what does the k=1 actually mean? does it means that he will have at least on g1? $\endgroup$ – Lothre1 Jan 5 '14 at 20:27
  • $\begingroup$ $k$ is a "dummy" variable, like a loop index in programming, going from 1 to $n$. $\endgroup$ – hardmath Jan 5 '14 at 20:28
  • $\begingroup$ ah ok. But how can i integrate the hole formula? Because i don't have math background and what confuses me is the way we represent those operations $\endgroup$ – Lothre1 Jan 5 '14 at 20:30

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