+118608 Hi Dragon Slayer,
I see you have found a new method for making 'legitimate' posts. I hope that you learn something from it. 
Anyway, I have learned something - I have never written this in LaTex before!
By the way, LaTex is pronounced LayTec
$$\\\sum \quad
\mbox{This is the Greek letter 'capital sigma' and it means 'sum of'}\\\\
\sum\limits_{n=1}^5\;(n^2-1)\\\\
=(1^2-1)+(2^2-1)+(3^2-1)+(4^2-1)+(5^2-1)\\\\
=1+3+8+15+24\\\\
=51\\\\
\mbox{You see how I have replaced the n with the numbers from 1 to 5 and added them all together?}$$
Oh sorry, I didn't even notice the original question. This was a puzzle question.
I think that Aziz answered it a day or 2 ago. I think it was Aziz, maybe it was Ninja but I don't think so. 
It was a really good answer too. 
+118608 Hi Dragon Slayer,
I see you have found a new method for making 'legitimate' posts. I hope that you learn something from it.
Anyway, I have learned something - I have never written this in LaTex before!
By the way, LaTex is pronounced LayTec
$$\\\sum \quad
\mbox{This is the Greek letter 'capital sigma' and it means 'sum of'}\\\\
\sum\limits_{n=1}^5\;(n^2-1)\\\\
=(1^2-1)+(2^2-1)+(3^2-1)+(4^2-1)+(5^2-1)\\\\
=1+3+8+15+24\\\\
=51\\\\
\mbox{You see how I have replaced the n with the numbers from 1 to 5 and added them all together?}$$
Oh sorry, I didn't even notice the original question. This was a puzzle question.
I think that Aziz answered it a day or 2 ago. I think it was Aziz, maybe it was Ninja but I don't think so.
It was a really good answer too.
