MathsGod1

I didn't even notice thanks i will use \\\\ and i will put in \sqrt

$$\\\\\left(z-\frac{2}{\sqrt{z}}\right)\\\\The\:general\:term\:is\\\\
=(9Cr)(z)(9-r)\left(\frac{z^9}{z^r}\right)\left(\frac{(-2)^r}{(\sqrt{z}^r}\right)\\\\
=(9Cr)\left(\frac{z^9*(-2)^r}{z^r*(\sqrt{z}^r)}\right)\\\\
=(9Cr)\left(\frac{*(-2)^r}{z^(2r/2)*(z)^(r/2)}\right)\\\\
=(9Cr)\left(\frac{z^{((18-2r-r)/2)}*(-2)^r}{1}\right)\\\\
=(9Cr)(z^{((18-3r)/2)}*(-2)^r)
\\\\The\:constant\:term\:will\:be\:when\\\\
(18-3r)/2=0\\18-3r=0\\r=6$$

What to do next.....Hmmmm...

I dont think my math knowledge has exceeded far enough to think of a new topic :)

I agree!

Thank you Melody!  :D    :D     :D

aha Melody,  I've noticed Heureka's LaTeX ever sinced i layed my eyes on this website and LaTeX.

The LaTeX is amazing but i think i should take it slow before i get to your level...Really slow :D

I'm on Nauseated side...This Titanium army has gone too far. It's even changing this sites name.

$$\\\left(z-\dfrac{2}{\sqrt{z}}\right)\\\\The\:general\:term\:is\\\\
=(9Cr)(z)(9-r)\left(\dfrac{z^9}{z^r}\right)\left(\dfrac{(-2)^r}{(\sqrt{z}^r}\right)\\
=(9Cr)\left(\dfrac{z^9*(-2)^r}{z^r*(sqrt{z}^r)}\right)\\
=(9Cr)\left(\dfrac{*(-2)^r}{z^(2r/2)*(z)^(r/2)}\right)\\
=(9Cr)\left(\dfrac{z^{((18-2r-r)/2)}*(-2)^r}{1}\right)\\
=(9Cr)(z^{((18-3r)/2)}*(-2)^r)
\\The\:constant\:term\:will\:be\:when\\
(18-3r)/2=0\\18-3r=0\\r=6$$

Is there a way to leave larger gaps it seems so cramped.

I'll try :))

I'll start it by continuing off from last time.

$$\\\left(z-\frac{2}{\sqrt{z}}\right)\\\\The\:general\:term\:is\\\\
=(9Cr)(z)(9-r)\left(\frac{z^9}{z^r}\right)\left(\frac{(-2)^r}{(\sqrt{z}^r}\right)\\
=(9Cr)\left(\frac{z^9*(-2)^r}{z^r*(sqrt{z}^r)}\right)\\
=(9Cr)\left(\frac{*(-2)^r}{z^(2r/2)*(z)^(r/2)}\right)\\
=(9Cr)\left(\frac{z^{((18-2r-r)/2)}*(-2)^r}{1}\right)\\
=(9Cr)(z^{((18-3r)/2)}*(-2)^r)
\\The\:constant\:term\:will\:be\:when\\
(18-3r)/2=0\\18-3r=0\\r=6$$

This is the entire coding correct...

Hopefully it shows...My laptop might not be able to take in all this LaTeX.

\\\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(\frac{z^{((18-2r-r)/2)}*(-2)^r}{1}\right)\\
=(9Cr)(z^{((18-3r)/2)}*(-2)^r)
\\The\:constant\:term\:will\:be\:when\\
(18-3r)/2=0\\18-3r=0\\r=6

Hahaha!

Lol, if I'm the god then I can't describe what Melody is, she taught me this!

$$\\\left(z-\frac{2}{\sqrt{z}}\right)\\\\The\:general\:term\:is\\\\
=(9Cr)(z)(9-r)\left(\frac{z^9}{z^r}\right)\left(\frac{(-2)^r}{(\sqrt{z}^r}\right)\\
=(9Cr)\left(\frac{z^9*(-2)^r}{z^r*(sqrt{z}^r)}\right)\\
=(9Cr)\left(\frac{*(-2)^r}{z^(2r/2)*(z)^(r/2)}\right)\\
=(9Cr)\left(\frac{z^{((18-2r-r)/2)}*(-2)^r}{1}\right)\\
=(9Cr)(z^{((18-3r)/2)}*(-2)^r)
\\The\:constant\:term\:will\:be\:when\\
(18-3r)/2=0\\18-3r=0\\r=6$$

This is the entire coding correct...

Hopefully it shows...My laptop might not be able to take in all this LaTeX.

\\\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(\frac{z^{((18-2r-r)/2)}*(-2)^r}{1}\right)\\
=(9Cr)(z^{((18-3r)/2)}*(-2)^r)
\\The\:constant\:term\:will\:be\:when\\
(18-3r)/2=0\\18-3r=0\\r=6

The code didn't display...

$$\\=(9Cr)\left(\frac{z^{((18-2r-r)/2)}*(-2)^r}{1}\right)\\
=(9Cr)(z^{((18-3r)/2)}*(-2)^r)
\\The\:constant\:term\:will\:be\:when\\
(18-3r)/2=0\\18-3r=0\\r=6$$

Here's the code Melody.

\\=(9Cr)\left(\frac{z^{((18-2r-r)/2)}*(-2)^r}{1}\right)\\
=(9Cr)(z^{((18-3r)/2)}*(-2)^r)
\\The\:constant\:term\:will\:be\:when\\
(18-3r)/2=0\\18-3r=0\\r=6