# Help with proof of Arithmetic mean

• Feb 13th 2008, 09:24 AM
janvdl
Help with proof of Arithmetic mean
$x = \frac{1}{n} \sum^{n}_{i = 1} x_{i}$

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

$x = \frac{1}{n} \left( \sum^{n}_{i = 1} x_{i} - na \right) + a$ <- 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. :D
• Feb 13th 2008, 09:36 AM
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}
$
• Feb 13th 2008, 09:39 AM
janvdl
Quote:

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?
• Feb 13th 2008, 09:45 AM
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)
$
• Feb 13th 2008, 09:52 AM
janvdl
Quote:

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. (Handshake) (Clapping)