LaTeX array with correlation coefficient formula not displaying

[bomba923]

What's wrong with my code? Why would LaTex show my work?

$$\begin{array}{l}<br /> r = \frac{1}{{k - 1}}\sum\limits_{n = 1}^k {\left[ {\left( {\frac{{x_n - \bar x}}{{S_x }}} \right)\left( {\frac{{y_n - \bar y}}{{S_y }}} \right)} \right]} = \left( {\frac{1}{{k - 1}}} \right)\sum\limits_{n = 1}^k {\left[ {\left( {\frac{{x_n - \frac{{\sum\limits_{n = 1}^k {x_n } }}{k}}}{{\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {x_n - \frac{{\sum\limits_{n = 1}^k {x_n } }}{k}} \right)^2 } }}{{k - 1}}} }}} \right)\left( {\frac{{y_n - \frac{{\sum\limits_{n = 1}^k {y_n } }}{k}}}{{\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {y_n - \frac{{\sum\limits_{n = 1}^k {y_n } }}{k}} \right)^2 } }}{{k - 1}}} }}} \right)} \right]} \Rightarrow \left( \begin{array}{l}<br /> {\rm multiply both fractions} \\ <br /> {\rm by }\frac{{\left( {{\raise0.7ex\hbox{1} \!\mathord{\left/<br /> {\vphantom {1 k}}\right.\kern-\nulldelimiterspace}<br /> \!\lower0.7ex\hbox{k}}} \right)}}{{\left( {{\raise0.7ex\hbox{1} \!\mathord{\left/<br /> {\vphantom {1 k}}\right.\kern-\nulldelimiterspace}<br /> \!\lower0.7ex\hbox{k}}} \right)}} = \frac{{k^{ - 1} }}{{k^{ - 1} }} = \frac{k}{k} = 1 \\ <br /> \end{array} \right) \Rightarrow \\ <br /> \left( {\frac{1}{{k - 1}}} \right)\sum\limits_{n = 1}^k {\left[ {\left( {\frac{{x_n k - \sum\limits_{n = 1}^k {x_n } }}{{k\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {x_n - \frac{{\sum\limits_{n = 1}^k {x_n } }}{k}} \right)^2 } }}{{k - 1}}} }}} \right)\left( {\frac{{y_n k - \sum\limits_{n = 1}^k {y_n } }}{{k\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {y_n - \frac{{\sum\limits_{n = 1}^k {y_n } }}{k}} \right)^2 } }}{{k - 1}}} }}} \right)} \right]} \Rightarrow \left( \begin{array}{l}<br /> {\rm rewrite the bottoms sums over} \\ <br /> {\rm a common denominator,}\;\left( k \right) \\ <br /> \end{array} \right) \Rightarrow \left( {\frac{1}{{k - 1}}} \right)\sum\limits_{n = 1}^k {\left[ {\left( {\frac{{x_n k - \sum\limits_{n = 1}^k {x_n } }}{{k\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {\frac{{x_n k - \sum\limits_{n = 1}^k {x_n } }}{k}} \right)^2 } }}{{k - 1}}} }}} \right)\left( {\frac{{y_n k - \sum\limits_{n = 1}^k {y_n } }}{{k\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {\frac{{y_n k - \sum\limits_{n = 1}^k {y_n } }}{k}} \right)^2 } }}{{k - 1}}} }}} \right)} \right]} \\ <br /> \Rightarrow \left( \begin{array}{l}<br /> {\rm Expand the bottom sums to factor } \\ <br /> {\rm out }\left( k \right).{\rm Because we have a sum of } \\ <br /> squares{\rm , we will factor out a }\left( {k^2 } \right) \\ <br /> \end{array} \right) \Rightarrow \left( {\frac{1}{{k - 1}}} \right)\sum\limits_{n = 1}^k {\left[ {\left( {\frac{{x_n k - \sum\limits_{n = 1}^k {x_n } }}{{k\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)^2 } }}{{k^2 \left( {k - 1} \right)}}} }}} \right)\left( {\frac{{y_n k - \sum\limits_{n = 1}^k {y_n } }}{{k\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)^2 } }}{{k^2 \left( {k - 1} \right)}}} }}} \right)} \right]} \Rightarrow \left( \begin{array}{l}<br /> {\rm From this, we can solve the square root of} \\ <br /> \;\left( {k^2 } \right)\;{\rm to place this value outside the radical } \\ <br /> \end{array} \right) \Rightarrow \\ <br /> \Rightarrow \left( {\frac{1}{{k - 1}}} \right)\sum\limits_{n = 1}^k {\left[ {\left( {\frac{{x_n k - \sum\limits_{n = 1}^k {x_n } }}{{\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)^2 } }}{{\left( {k - 1} \right)}}} }}} \right)\left( {\frac{{y_n k - \sum\limits_{n = 1}^k {y_n } }}{{\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)^2 } }}{{\left( {k - 1} \right)}}} }}} \right)} \right]} \Rightarrow \left( \begin{array}{l}<br /> {\rm The }\left( k \right){\rm values will cancel } \\ <br /> {\rm as we multiply:}\;k \cdot \sqrt {\frac{1}{{k^2 }}} = 1 \\ <br /> \end{array} \right) \Rightarrow \left( {\frac{1}{{k - 1}}} \right)\sum\limits_{n = 1}^k {\left[ {\left( {\frac{{x_n k - \sum\limits_{n = 1}^k {x_n } }}{{\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)^2 } }}{{\left( {k - 1} \right)}}} }}} \right)\left( {\frac{{y_n k - \sum\limits_{n = 1}^k {y_n } }}{{\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)^2 } }}{{\left( {k - 1} \right)}}} }}} \right)} \right]} \Rightarrow \\ <br /> \Rightarrow \left( \begin{array}{l}<br /> {\rm We express the bottom radical as a} \\ <br /> {\rm a radical numerator divided by }\sqrt {k - 1} ; \\ <br /> {\rm therefore, we can place }\sqrt {k - 1} {\rm in the } \\ <br /> {\rm numerator of the entire equation } \\ <br /> \end{array} \right) \Rightarrow \left( {\frac{1}{{k - 1}}} \right)\sum\limits_{n = 1}^k {\left\{ {\left[ {\frac{{\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)\sqrt {k - 1} }}{{\sqrt {\sum\limits_{n = 1}^k {\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)^2 } } }}} \right]\left[ {\frac{{\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)\sqrt {k - 1} }}{{\sqrt {\sum\limits_{n = 1}^k {\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)^2 } } }}} \right]} \right\}} \Rightarrow \left( \begin{array}{l}<br /> {\rm We can multiply both z - score expressions} \\ <br /> {\rm to simplify the sum and remove }\sqrt {k - 1} \\ <br /> \end{array} \right) \Rightarrow \\ <br /> \Rightarrow \left( {\frac{1}{{k - 1}}} \right)\sum\limits_{n = 1}^k {\left[ {\frac{{\left( {k - 1} \right)\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)}}{{\left( {\sqrt {\sum\limits_{n = 1}^k {\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)^2 } } } \right)\left( {\sqrt {\sum\limits_{n = 1}^k {\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)^2 } } } \right)}}} \right]} = \sum\limits_{n = 1}^k {\left\{ {\frac{{\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)}}{{\sqrt {\left[ {\sum\limits_{n = 1}^k {\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)^2 } } \right]\left[ {\sum\limits_{n = 1}^k {\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)^2 } } \right]} }}} \right\}} = = \frac{{\sum\limits_{n = 1}^k {\left[ {\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)} \right]} }}{{\sqrt {\left[ {\sum\limits_{n = 1}^k {\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)^2 } } \right]\left[ {\sum\limits_{n = 1}^k {\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)^2 } } \right]} }} \Rightarrow \\ <br /> \Rightarrow \left( {{\rm Substitute}\left\{ \begin{array}{l}<br /> a = \sum\limits_{n = 1}^k {x_n } \\ <br /> b = \sum\limits_{n = 1}^k {y_n } \\ <br /> \end{array} \right\}} \right) \Rightarrow \frac{{\sum\limits_{n = 1}^k {\left[ {\left( {x_n y_n } \right)k^2 - k\left( {y_n a + x_n b} \right) + ab} \right]} }}{{\sqrt {\left[ {\sum\limits_{n = 1}^k {\left( {x_n^2 k^2 - 2ax_n k + a^2 } \right)} } \right]\left[ {\sum\limits_{n = 1}^k {y_n^2 k^2 - 2by_n k + b^2 } } \right]} }} = \Rightarrow \left( \begin{array}{l}<br /> {\rm We can take out}\left( {ab} \right){\rm from the } \\ <br /> {\rm numerator and the}\left( {a^2 ,b^2 } \right){\rm from the } \\ <br /> {\rm denominator as}\left( {kab} \right){\rm and}\left( {a^2 k,b^2 k} \right) \\ <br /> \end{array} \right) \Rightarrow \frac{{kab + \sum\limits_{n = 1}^k {\left[ {\left( {x_n y_n } \right)k^2 - k\left( {y_n a + x_n b} \right)} \right]} }}{{\sqrt {\left[ {a^2 k + \sum\limits_{n = 1}^k {\left( {x_n^2 k^2 - 2ax_n k} \right)} } \right]\left[ {b^2 k + \sum\limits_{n = 1}^k {\left( {y_n^2 k^2 - 2by_n k} \right)} } \right]} }} \Rightarrow \\ <br /> \Rightarrow \left( \begin{array}{l}<br /> {\rm Then, we can factor out the} \\ <br /> \left( k \right){\rm from the sum expressions} \\ <br /> \end{array} \right) \Rightarrow = \frac{{kab + k\sum\limits_{n = 1}^k {\left[ {\left( {x_n y_n } \right)k - \left( {y_n a + x_n b} \right)} \right]} }}{{\sqrt {\left[ {a^2 k + k\sum\limits_{n = 1}^k {\left( {x_n^2 k - 2x_n a} \right)} } \right]\left[ {b^2 k + k\sum\limits_{n = 1}^k {\left( {y_n^2 k - 2y_n b} \right)} } \right]} }} = = \frac{{k\left\{ {ab + \sum\limits_{n = 1}^k {\left[ {\left( {x_n y_n } \right)k - \left( {y_n a + x_n b} \right)} \right]} } \right\}}}{{\sqrt {k^2 \left[ {a^2 + \sum\limits_{n = 1}^k {\left( {x_n^2 k - 2x_n a} \right)} } \right]\left[ {b^2 + \sum\limits_{n = 1}^k {\left( {y_n^2 k - 2y_n b} \right)} } \right]} }} \Rightarrow \left( \begin{array}{l}<br /> {\rm The}\left( {k^2 } \right){\rm in the denominator} \\ <br /> {\rm radical can be taken out and} \\ <br /> {\rm will cancel with the }\left( k \right) \\ <br /> {\rm in the equation numerator} \\ <br /> \end{array} \right) \\ <br /> = \frac{{ab + \sum\limits_{n = 1}^k {\left( {x_n y_n k - x_n b - y_n a} \right)} }}{{\sqrt {\left[ {a^2 + \sum\limits_{n = 1}^k {\left( {x_n^2 k - 2x_n a} \right)} } \right]\left[ {b^2 + \sum\limits_{n = 1}^k {\left( {y_n^2 k - 2y_n b} \right)} } \right]} }} = \frac{{ab + \sum\limits_{n = 1}^k {\left( {x_n y_n k - x_n b - y_n a} \right)} }}{{\sqrt {\left\{ {a^2 + \sum\limits_{n = 1}^k {\left[ {x_n \left( {x_n - 2a} \right)} \right]} } \right\}\left\{ {b^2 + \sum\limits_{n = 1}^k {\left[ {y_n \left( {y_n - 2b} \right)} \right]} } \right\}} }} \Rightarrow \left( \begin{array}{l}<br /> {\rm Replace variables }\left( {a,b} \right) \\ <br /> {\rm with their original assignments} \\ <br /> \end{array} \right) \Rightarrow \\ <br /> \frac{{\left( {\sum\limits_{n = 1}^k {x_n } } \right)\left( {\sum\limits_{n = 1}^k {y_n } } \right) + \sum\limits_{n = 1}^k {\left( {x_n y_n k - x_n \sum\limits_{n = 1}^k {y_n } - y_n \sum\limits_{n = 1}^k {x_n } } \right)} }}{{\sqrt {\left[ {\left( {\sum\limits_{n = 1}^k {x_n } } \right)^2 + \sum\limits_{n = 1}^k {\left( {x_n^2 k - 2x_n \sum\limits_{n = 1}^k {x_n } } \right)} } \right]\left[ {\left( {\sum\limits_{n = 1}^k {y_n } } \right)^2 + \sum\limits_{n = 1}^k {\left( {y_n^2 k - 2y_n \sum\limits_{n = 1}^k {y_n } } \right)} } \right]} }} = \frac{{\left( {\sum\limits_{n = 1}^k {x_n } } \right)\left( {\sum\limits_{n = 1}^k {y_n } } \right) + \sum\limits_{n = 1}^k {\left( {x_n y_n k - x_n \sum\limits_{n = 1}^k {y_n } - y_n \sum\limits_{n = 1}^k {x_n } } \right)} }}{{\sqrt {\left\{ {\left( {\sum\limits_{n = 1}^k {x_n } } \right)^2 + \sum\limits_{n = 1}^k {\left[ {x_n \left( {x_n - 2\sum\limits_{n = 1}^k {x_n } } \right)} \right]} } \right\}\left\{ {\left( {\sum\limits_{n = 1}^k {y_n } } \right)^2 + \sum\limits_{n = 1}^k {\left[ {y_n \left( {y_n - 2\sum\limits_{n = 1}^k {y_n } } \right)} \right]} } \right\}} }} \\ <br /> \end{array}$$

What does it mean;--what's "invalid" here?

 

[bomba923]

Hmm---I think I fixed it; how do you know when to press enter and start a new command line, so you can see my WHOLE work? (because browser's don't scroll that far!)
$$\begin{gathered}<br /> r = \frac{1}<br /> {{k - 1}}\sum\limits_{n = 1}^k {\left[ {\left( {\frac{{x_n - \bar x}}<br /> {{S_x }}} \right)\left( {\frac{{y_n - \bar y}}<br /> {{S_y }}} \right)} \right]} = \left( {\frac{1}<br /> {{k - 1}}} \right)\sum\limits_{n = 1}^k {\left[ {\left( {\frac{{x_n - \frac{{\sum\limits_{n = 1}^k {x_n } }}<br /> {k}}}<br /> {{\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {x_n - \frac{{\sum\limits_{n = 1}^k {x_n } }}<br /> {k}} \right)^2 } }}<br /> {{k - 1}}} }}} \right)\left( {\frac{{y_n - \frac{{\sum\limits_{n = 1}^k {y_n } }}<br /> {k}}}<br /> {{\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {y_n - \frac{{\sum\limits_{n = 1}^k {y_n } }}<br /> {k}} \right)^2 } }}<br /> {{k - 1}}} }}} \right)} \right]} \Rightarrow \left( \begin{gathered}<br /> \text{multiply both fractions} \hfill \\<br /> \text{by }\frac{{\left( {{\raise0.7ex\hbox{1} \!\mathord{\left/<br /> {\vphantom {1 k}}\right.\kern-\nulldelimiterspace}<br /> \!\lower0.7ex\hbox{k}}} \right)}}<br /> {{\left( {{\raise0.7ex\hbox{1} \!\mathord{\left/<br /> {\vphantom {1 k}}\right.\kern-\nulldelimiterspace}<br /> \!\lower0.7ex\hbox{k}}} \right)}} = \frac{{k^{ - 1} }}<br /> {{k^{ - 1} }} = \frac{k}<br /> {k} = 1 \hfill \\ <br /> \end{gathered} \right) \Rightarrow \hfill \\<br /> \left( {\frac{1}<br /> {{k - 1}}} \right)\sum\limits_{n = 1}^k {\left[ {\left( {\frac{{x_n k - \sum\limits_{n = 1}^k {x_n } }}<br /> {{k\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {x_n - \frac{{\sum\limits_{n = 1}^k {x_n } }}<br /> {k}} \right)^2 } }}<br /> {{k - 1}}} }}} \right)\left( {\frac{{y_n k - \sum\limits_{n = 1}^k {y_n } }}<br /> {{k\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {y_n - \frac{{\sum\limits_{n = 1}^k {y_n } }}<br /> {k}} \right)^2 } }}<br /> {{k - 1}}} }}} \right)} \right]} \Rightarrow \left( \begin{gathered}<br /> \text{rewrite the bottoms sums over} \hfill \\<br /> \text{a common denominator,}\;\left( k \right) \hfill \\ <br /> \end{gathered} \right) \Rightarrow \left( {\frac{1}<br /> {{k - 1}}} \right)\sum\limits_{n = 1}^k {\left[ {\left( {\frac{{x_n k - \sum\limits_{n = 1}^k {x_n } }}<br /> {{k\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {\frac{{x_n k - \sum\limits_{n = 1}^k {x_n } }}<br /> {k}} \right)^2 } }}<br /> {{k - 1}}} }}} \right)\left( {\frac{{y_n k - \sum\limits_{n = 1}^k {y_n } }}<br /> {{k\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {\frac{{y_n k - \sum\limits_{n = 1}^k {y_n } }}<br /> {k}} \right)^2 } }}<br /> {{k - 1}}} }}} \right)} \right]} \hfill \\<br /> \Rightarrow \left( \begin{gathered}<br /> \text{Expand the bottom sums to factor } \hfill \\<br /> \text{out }\left( k \right).\text{ Because we have a sum of } \hfill \\<br /> squares\text{, we will factor out a }\left( {k^2 } \right) \hfill \\ <br /> \end{gathered} \right) \Rightarrow \left( {\frac{1}<br /> {{k - 1}}} \right)\sum\limits_{n = 1}^k {\left[ {\left( {\frac{{x_n k - \sum\limits_{n = 1}^k {x_n } }}<br /> {{k\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)^2 } }}<br /> {{k^2 \left( {k - 1} \right)}}} }}} \right)\left( {\frac{{y_n k - \sum\limits_{n = 1}^k {y_n } }}<br /> {{k\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)^2 } }}<br /> {{k^2 \left( {k - 1} \right)}}} }}} \right)} \right]} \Rightarrow \left( \begin{gathered}<br /> \text{From this, we can solve the square root of} \hfill \\<br /> \;\left( {k^2 } \right)\;\text{to place this value outside the radical } \hfill \\ <br /> \end{gathered} \right) \Rightarrow \hfill \\<br /> \Rightarrow \left( {\frac{1}<br /> {{k - 1}}} \right)\sum\limits_{n = 1}^k {\left[ {\left( {\frac{{x_n k - \sum\limits_{n = 1}^k {x_n } }}<br /> {{\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)^2 } }}<br /> {{\left( {k - 1} \right)}}} }}} \right)\left( {\frac{{y_n k - \sum\limits_{n = 1}^k {y_n } }}<br /> {{\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)^2 } }}<br /> {{\left( {k - 1} \right)}}} }}} \right)} \right]} \Rightarrow \left( \begin{gathered}<br /> \text{The }\left( k \right)\text{values will cancel } \hfill \\<br /> \text{as we multiply:}\;k \cdot \sqrt {\frac{1}<br /> {{k^2 }}} = 1 \hfill \\ <br /> \end{gathered} \right) \Rightarrow \left( {\frac{1}<br /> {{k - 1}}} \right)\sum\limits_{n = 1}^k {\left[ {\left( {\frac{{x_n k - \sum\limits_{n = 1}^k {x_n } }}<br /> {{\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)^2 } }}<br /> {{\left( {k - 1} \right)}}} }}} \right)\left( {\frac{{y_n k - \sum\limits_{n = 1}^k {y_n } }}<br /> {{\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)^2 } }}<br /> {{\left( {k - 1} \right)}}} }}} \right)} \right]} \Rightarrow \hfill \\<br /> \Rightarrow \left( \begin{gathered}<br /> \text{We express the bottom radical as a} \hfill \\<br /> \text{a radical numerator divided by }\sqrt {k - 1} ; \hfill \\<br /> \text{therefore, we can place }\sqrt {k - 1} \text{ in the } \hfill \\<br /> \text{ numerator of the entire equation } \hfill \\ <br /> \end{gathered} \right) \Rightarrow \left( {\frac{1}<br /> {{k - 1}}} \right)\sum\limits_{n = 1}^k {\left\{ {\left[ {\frac{{\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)\sqrt {k - 1} }}<br /> {{\sqrt {\sum\limits_{n = 1}^k {\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)^2 } } }}} \right]\left[ {\frac{{\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)\sqrt {k - 1} }}<br /> {{\sqrt {\sum\limits_{n = 1}^k {\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)^2 } } }}} \right]} \right\}} \Rightarrow \left( \begin{gathered}<br /> \text{We can multiply both z - score expressions} \hfill \\<br /> \text{to simplify the sum and remove }\sqrt {k - 1} \hfill \\ <br /> \end{gathered} \right) \Rightarrow \hfill \\<br /> \Rightarrow \left( {\frac{1}<br /> {{k - 1}}} \right)\sum\limits_{n = 1}^k {\left[ {\frac{{\left( {k - 1} \right)\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)}}<br /> {{\left( {\sqrt {\sum\limits_{n = 1}^k {\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)^2 } } } \right)\left( {\sqrt {\sum\limits_{n = 1}^k {\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)^2 } } } \right)}}} \right]} = \sum\limits_{n = 1}^k {\left\{ {\frac{{\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)}}<br /> {{\sqrt {\left[ {\sum\limits_{n = 1}^k {\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)^2 } } \right]\left[ {\sum\limits_{n = 1}^k {\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)^2 } } \right]} }}} \right\}} = = \frac{{\sum\limits_{n = 1}^k {\left[ {\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)} \right]} }}<br /> {{\sqrt {\left[ {\sum\limits_{n = 1}^k {\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)^2 } } \right]\left[ {\sum\limits_{n = 1}^k {\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)^2 } } \right]} }} \Rightarrow \hfill \\<br /> \Rightarrow \left( {\text{Substitute}\left\{ \begin{gathered}<br /> a = \sum\limits_{n = 1}^k {x_n } \hfill \\<br /> b = \sum\limits_{n = 1}^k {y_n } \hfill \\ <br /> \end{gathered} \right\}} \right) \Rightarrow \frac{{\sum\limits_{n = 1}^k {\left[ {\left( {x_n y_n } \right)k^2 - k\left( {y_n a + x_n b} \right) + ab} \right]} }}<br /> {{\sqrt {\left[ {\sum\limits_{n = 1}^k {\left( {x_n^2 k^2 - 2ax_n k + a^2 } \right)} } \right]\left[ {\sum\limits_{n = 1}^k {y_n^2 k^2 - 2by_n k + b^2 } } \right]} }} = \Rightarrow \left( \begin{gathered}<br /> \text{We can take out}\left( {ab} \right)\text{from the } \hfill \\<br /> \text{numerator and the}\left( {a^2 ,b^2 } \right)\text{from the } \hfill \\<br /> \text{denominator as}\left( {kab} \right)\text{and}\left( {a^2 k,b^2 k} \right) \hfill \\ <br /> \end{gathered} \right) \Rightarrow \frac{{kab + \sum\limits_{n = 1}^k {\left[ {\left( {x_n y_n } \right)k^2 - k\left( {y_n a + x_n b} \right)} \right]} }}<br /> {{\sqrt {\left[ {a^2 k + \sum\limits_{n = 1}^k {\left( {x_n^2 k^2 - 2ax_n k} \right)} } \right]\left[ {b^2 k + \sum\limits_{n = 1}^k {\left( {y_n^2 k^2 - 2by_n k} \right)} } \right]} }} \Rightarrow \hfill \\<br /> \Rightarrow \left( \begin{gathered}<br /> \text{Then, we can factor out the} \hfill \\<br /> \left( k \right)\text{from the sum expressions} \hfill \\ <br /> \end{gathered} \right) \Rightarrow = \frac{{kab + k\sum\limits_{n = 1}^k {\left[ {\left( {x_n y_n } \right)k - \left( {y_n a + x_n b} \right)} \right]} }}<br /> {{\sqrt {\left[ {a^2 k + k\sum\limits_{n = 1}^k {\left( {x_n^2 k - 2x_n a} \right)} } \right]\left[ {b^2 k + k\sum\limits_{n = 1}^k {\left( {y_n^2 k - 2y_n b} \right)} } \right]} }} = = \frac{{k\left\{ {ab + \sum\limits_{n = 1}^k {\left[ {\left( {x_n y_n } \right)k - \left( {y_n a + x_n b} \right)} \right]} } \right\}}}<br /> {{\sqrt {k^2 \left[ {a^2 + \sum\limits_{n = 1}^k {\left( {x_n^2 k - 2x_n a} \right)} } \right]\left[ {b^2 + \sum\limits_{n = 1}^k {\left( {y_n^2 k - 2y_n b} \right)} } \right]} }} \Rightarrow \left( \begin{gathered}<br /> \text{The}\left( {k^2 } \right)\text{in the denominator} \hfill \\<br /> \text{radical can be taken out and} \hfill \\<br /> \text{will cancel with the }\left( k \right) \hfill \\<br /> \text{in the equation numerator} \hfill \\ <br /> \end{gathered} \right) \hfill \\<br /> = \frac{{ab + \sum\limits_{n = 1}^k {\left( {x_n y_n k - x_n b - y_n a} \right)} }}<br /> {{\sqrt {\left[ {a^2 + \sum\limits_{n = 1}^k {\left( {x_n^2 k - 2x_n a} \right)} } \right]\left[ {b^2 + \sum\limits_{n = 1}^k {\left( {y_n^2 k - 2y_n b} \right)} } \right]} }} = \frac{{ab + \sum\limits_{n = 1}^k {\left( {x_n y_n k - x_n b - y_n a} \right)} }}<br /> {{\sqrt {\left\{ {a^2 + \sum\limits_{n = 1}^k {\left[ {x_n \left( {x_n - 2a} \right)} \right]} } \right\}\left\{ {b^2 + \sum\limits_{n = 1}^k {\left[ {y_n \left( {y_n - 2b} \right)} \right]} } \right\}} }} \Rightarrow \left( \begin{gathered}<br /> \text{Replace variables }\left( {a,b} \right) \hfill \\<br /> \text{with their original assignments} \hfill \\ <br /> \end{gathered} \right) \Rightarrow \hfill \\<br /> \frac{{\left( {\sum\limits_{n = 1}^k {x_n } } \right)\left( {\sum\limits_{n = 1}^k {y_n } } \right) + \sum\limits_{n = 1}^k {\left( {x_n y_n k - x_n \sum\limits_{n = 1}^k {y_n } - y_n \sum\limits_{n = 1}^k {x_n } } \right)} }}<br /> {{\sqrt {\left[ {\left( {\sum\limits_{n = 1}^k {x_n } } \right)^2 + \sum\limits_{n = 1}^k {\left( {x_n^2 k - 2x_n \sum\limits_{n = 1}^k {x_n } } \right)} } \right]\left[ {\left( {\sum\limits_{n = 1}^k {y_n } } \right)^2 + \sum\limits_{n = 1}^k {\left( {y_n^2 k - 2y_n \sum\limits_{n = 1}^k {y_n } } \right)} } \right]} }} = \frac{{\left( {\sum\limits_{n = 1}^k {x_n } } \right)\left( {\sum\limits_{n = 1}^k {y_n } } \right) + \sum\limits_{n = 1}^k {\left( {x_n y_n k - x_n \sum\limits_{n = 1}^k {y_n } - y_n \sum\limits_{n = 1}^k {x_n } } \right)} }}<br /> {{\sqrt {\left\{ {\left( {\sum\limits_{n = 1}^k {x_n } } \right)^2 + \sum\limits_{n = 1}^k {\left[ {x_n \left( {x_n - 2\sum\limits_{n = 1}^k {x_n } } \right)} \right]} } \right\}\left\{ {\left( {\sum\limits_{n = 1}^k {y_n } } \right)^2 + \sum\limits_{n = 1}^k {\left[ {y_n \left( {y_n - 2\sum\limits_{n = 1}^k {y_n } } \right)} \right]} } \right\}} }} \hfill \\ <br /> \end{gathered}$$

 

[dextercioby]

1.It's in the wrong forum;
2.Don't abuse the latex code and mess up page layout.I don't want to change the # of pixels only to see your nonsense...
3.Use \\ or simply break the code (use [ tex ] & [ /tex ] tags more frequently.

Oh.4.Did i say that your post is a monstruosity which should be deleted?

Daniel.

 

[Evo]

Be nice!

But that is the largest I've seen.

bomba, I suggest you experiment on posts and use the "preview post" feature. Perhaps if you look at some posts by others that use LaTex and hit the "quote" button, as if you were going to reply, you will see how they did the formatting.

 

[dextercioby]

Sorry,Evo.I'm not (a) Saint : I overreact,sometimes...

Daniel.

P.S.The monstrusity part is not real,it's SURREAL