#2**+10 **

$$\\\left(z-\frac{2}{\sqrt{z}}\right)^9\\\\

$The general term is$\\\\

(9Cr)(z)^{(9-r)}\left(\frac{-2}{\sqrt{z}\right)^r}\\\\

=(9Cr)\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)}\right)\\\\

=(9Cr)\left(\dfrac{z^{18/2}*(-2)^r}{z^{(2r/2)}*(z)^{(r/2)}\right)}\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\\

S$so the constant term is $\\

=(9C6)*(-2)^6)\\

=64*9C6\\\\$$

$${\mathtt{64}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{9}}{!}}{{\mathtt{6}}{!}{\mathtt{\,\times\,}}({\mathtt{9}}{\mathtt{\,-\,}}{\mathtt{6}}){!}}}\right)} = {\mathtt{5\,376}}$$

You had better check that answer Mellie

Melody
May 7, 2015

#1**+9 **

We'll let WolframAlpha do the heavy lifting, here....

Looks like the constant term is 5376....

We could have actually determined this through the binomial expansion

The term would be C(9,6)z^3 [-2*z^(-1/2)]^6 = C(9,6)[-2]^6 = 5376 .....note that the z's "cancel"

CPhill
May 7, 2015

#2**+10 **

Best Answer

$$\\\left(z-\frac{2}{\sqrt{z}}\right)^9\\\\

$The general term is$\\\\

(9Cr)(z)^{(9-r)}\left(\frac{-2}{\sqrt{z}\right)^r}\\\\

=(9Cr)\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)}\right)\\\\

=(9Cr)\left(\dfrac{z^{18/2}*(-2)^r}{z^{(2r/2)}*(z)^{(r/2)}\right)}\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\\

S$so the constant term is $\\

=(9C6)*(-2)^6)\\

=64*9C6\\\\$$

$${\mathtt{64}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{9}}{!}}{{\mathtt{6}}{!}{\mathtt{\,\times\,}}({\mathtt{9}}{\mathtt{\,-\,}}{\mathtt{6}}){!}}}\right)} = {\mathtt{5\,376}}$$

You had better check that answer Mellie

Melody
May 7, 2015