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Find the constant term in the expansion of
\[\Big(z - \frac{2}{\sqrt{z}}\Big)^9.\]

Mellie  May 7, 2015

Best Answer 

 #2
avatar+93356 
+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
avatar+88898 
+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
avatar+93356 
+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
 #3
avatar+93356 
+4

WOW I got the same answer as CPhill   

Melody  May 7, 2015

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