Thread: Help with proof of Arithmetic mean

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

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

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

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

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

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.

Is this what you are missing?
$\displaystyle \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}
$

Originally Posted by Plato

Is this what you are missing?
$\displaystyle \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 $\displaystyle n $ of $\displaystyle na$ get factored out somehow?

Is this what you are missing?
$\displaystyle \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)
$

Originally Posted by Plato

Is this what you are missing?
$\displaystyle \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.