\\\left(z-\frac{2}{\sqrt{z}}\right)\\$$\\The\:general\:term\:is\\$$\\
=(9Cr)(z)(9-r)\left(\frac{z^9}{z^r}\right)\left(\frac{(-2)^r}{(\sqrt{z}^r}\right)$$\\$$
=(9Cr)\left(\frac{z^9*(-2)^r}{z^r*(sqrt{z}^r)}\right)$$\\$$
=(9Cr)\left(\frac{z^{18/2}*(-2)^r}{z^(2r/2)*(z)^(r/2)\right)$$\\$$
=(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
This was the code at the top of my head.
(It still buffers - crazy!)
The bit I'm confused in...
\\\left(z-\frac{2}{\sqrt{z}}\right)\\$$\\The\:general\:term\:is\\$$\\
=(9Cr)(z)(9-r)\left(\frac{z^9}{z^r}\right)\left(\frac{(-2)^r}{(\sqrt{z}^r}\right)$$\\$$
=(9Cr)\left(\frac{z^9*(-2)^r}{z^r*(sqrt{z}^r)}\right)$$\\$$
=(9Cr)\left(\frac{*(-2)^r}{z^(2r/2)*(z)^(r/2)\right)$$\\$$
=(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
The bold bit. Theres more of that...But why is does it come out normal.
$$z^{18/2}$$
When { } is taken out
$$z^18/2$$
I would just add { } but it affect my fractions...
EDIT: Ok, I'll do what you said.
It's been buffering for about 5 minutes i think it's because the code is wrong so its not responding but i don't know what's wrong.
http://i.imgur.com/lh7sCU8.png
