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# Find the constant term in the expansion of

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Find the constant term in the expansion of

Mellie  May 7, 2015

#2
+91436
+10

$$\\\left(z-\frac{2}{\sqrt{z}}\right)^9\\\\ The general term is\\\\ (9Cr)(z)^{(9-r)}\left(\frac{-2}{\sqrt{z}\right)^r}\\\\ =(9Cr)\left(\dfrac{z^9}{z^r}\right)\left(\dfrac{(-2)^r}{(\sqrt{z})^r\right)}\\\\ =(9Cr)\left(\dfrac{z^9*(-2)^r}{z^r*(\sqrt{z})^r\right)}\right)\\\\ =(9Cr)\left(\dfrac{z^{18/2}*(-2)^r}{z^{(2r/2)}*(z)^{(r/2)}\right)}\right)\\\\ =(9Cr)\left(\dfrac{z^{((18-2r-r)/2)}*(-2)^r}{1}\right)\\\\ =(9Cr)(z^{((18-3r)/2)}*(-2)^r)\\\\$$

$$\\The constant term will be when \\\\ (18-3r)/2=0\\ 18-3r=0\\ r=6\\ Sso the constant term is \\ =(9C6)*(-2)^6)\\ =64*9C6\\\\$$

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

Melody  May 7, 2015
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#1
+80913
+10

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
+91436
+10

$$\\\left(z-\frac{2}{\sqrt{z}}\right)^9\\\\ The general term is\\\\ (9Cr)(z)^{(9-r)}\left(\frac{-2}{\sqrt{z}\right)^r}\\\\ =(9Cr)\left(\dfrac{z^9}{z^r}\right)\left(\dfrac{(-2)^r}{(\sqrt{z})^r\right)}\\\\ =(9Cr)\left(\dfrac{z^9*(-2)^r}{z^r*(\sqrt{z})^r\right)}\right)\\\\ =(9Cr)\left(\dfrac{z^{18/2}*(-2)^r}{z^{(2r/2)}*(z)^{(r/2)}\right)}\right)\\\\ =(9Cr)\left(\dfrac{z^{((18-2r-r)/2)}*(-2)^r}{1}\right)\\\\ =(9Cr)(z^{((18-3r)/2)}*(-2)^r)\\\\$$

$$\\The constant term will be when \\\\ (18-3r)/2=0\\ 18-3r=0\\ r=6\\ Sso the constant term is \\ =(9C6)*(-2)^6)\\ =64*9C6\\\\$$

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