Hi MathsGod1 
Lets look at this one line at a time
LINE 1
\\\left(z-\frac{2}{\sqrt{z}}\right)\\$$\\The\:general\:term\:is\\$$\\
$$\\\left(z-\frac{2}{\sqrt{z}}\right)\\$$\\The\:general\:term\:is\\$$\\$$
Well the output is good but Iam wondering what those 2 double dollar signs are for.
I doubt that they do anything
LINE 2
=(9Cr)(z)(9-r)\left(\frac{z^9}{z^r}\right)\left(\frac{(-2)^r}{(\sqrt{z}^r}\right)$$\\$$
$$=(9Cr)(z)(9-r)\left(\frac{z^9}{z^r}\right)\left(\frac{(-2)^r}{(\sqrt{z}^r}\right)$$\\$$
That looks good but you are missing a little bracket :)
LINE 3
=(9Cr)\left(\frac{z^9*(-2)^r}{z^r*(sqrt{z}^r)}\right)$$\\$$
$$=(9Cr)\left(\frac{z^9*(-2)^r}{z^r*(sqrt{z}^r)}\right)$$\\$$$$
that is good - you are missing a backslash though. :)
LINE 4
=(9Cr)\left(\frac{z^{18/2}*(-2)^r}{z^(2r/2)*(z)^(r/2)\right)$$\\$$
That one doesn't work at all, I can see straight off that you are missing a }
Once you have made one fatal error, nothing will work, and that is a fatal error.
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A quick look at the rest and I can see that right is spelt incorrectly. I do not know if there are any other problems. but please try it without the double dollar signs - where did you get that idea from 
=(9Cr)\left(z^((18-2r-r))/2)*(-2)^r{1}\righr)$$\\$$
=(9Cr)(z^((12-3r)/2)*(-2)^r)$$\\$$
The\:constant\:term\:will\:be\:when$$\\$$
(18-3r)/2=0\\
18-3r=0\\
r=6
There you go - you try fixing this bit - I'll be back but I have to do some other work on the forum :)
