Help with proof of Arithmetic mean

**janvdl** · Feb 13th 2008, 09:24 AM

$x = \frac{1}{n} \sum^{n}_{i = 1} x_{i}$

$x = \frac{1}{n} \sum^{n}_{i = 1} x_{i} - a + a$

$x = \frac{1}{n} \left( \sum^{n}_{i = 1} x_{i} - na \right) + a$ <- What the heck happened with that $na$ thing? The transformation from the previous step, to this one, and the next step?

$x = \frac{1}{n} \left[ \sum^{n}_{i = 1} x_{i} - a \right] + a$

$x = \frac{ \sum^{n}_{i = 1} (x_{i} - a) }{n} + a$

By the way on the x on the left hand side there should be a little bar on top, indicating average.

Thanks for any help.

**Plato** · Feb 13th 2008, 09:36 AM

Is this what you are missing?
$\begin{gathered} \sum\limits_{k = 1}^n {\left( {x_k - a} \right)} = \left( {\sum\limits_{k = 1}^n {x_k } } \right) - na \hfill \\ \frac{1} {n}\left[ {\left( {\sum\limits_{k = 1}^n {x_k } } \right) - na} \right] = \frac{1} {n}\left( {\sum\limits_{k = 1}^n {x_k } } \right) - a \hfill \\ \end{gathered} $

**janvdl** · Feb 13th 2008, 09:39 AM

Originally Posted by Plato

Is this what you are missing?
$\begin{gathered} \sum\limits_{k = 1}^n {\left( {x_k - a} \right)} = \left( {\sum\limits_{k = 1}^n {x_k } } \right) - na \hfill \\ \frac{1} {n}\left[ {\left( {\sum\limits_{k = 1}^n {x_k } } \right) - na} \right] = \frac{1} {n}\left( {\sum\limits_{k = 1}^n {x_k } } \right) - a \hfill \\ \end{gathered} $

But didn't that $n$ of $na$ get factored out somehow?

**Plato** · Feb 13th 2008, 09:45 AM

Is this what you are missing?
$\frac{1} {n}\left[ {\left( {\sum\limits_{k = 1}^n {x_k } } \right) - na} \right] = \frac{1} {n}\left( {\sum\limits_{k = 1}^n {x_k } } \right) - \frac{1} {n}\left( {na} \right) $

**janvdl** · Feb 13th 2008, 09:52 AM

Originally Posted by Plato

Is this what you are missing?
$\frac{1} {n}\left[ {\left( {\sum\limits_{k = 1}^n {x_k } } \right) - na} \right] = \frac{1} {n}\left( {\sum\limits_{k = 1}^n {x_k } } \right) - \frac{1} {n}\left( {na} \right) $

There we go, that works for me. Thanks Plato.

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