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# Help!!

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Find the constant term in the expansion of $\Big(2z - \frac{1}{\sqrt{z}}\Big)^9.$

Oct 10, 2022

#1
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The general term is:

$$\displaystyle \binom{9}{n}\;(2z)^{(9-n)}(\frac{-1}{\sqrt z})^n\\~\\ =\displaystyle \binom{9}{n}\;(-1)^n(2)^{(9-n)}(\frac{1}{\sqrt z})^n*z^{(9-n)}\\~\\ =\displaystyle \binom{9}{n}\;(-1)^n(2)^{(9-n)}*z^{( \frac{-n+18-2n}{2} )}\\~\\ =\displaystyle \binom{9}{n}\;(-1)^n(2)^{(9-n)}*z^{( \frac{18-3n}{2} )}\\~\\ \qquad \text{You want the constant so}\\ \qquad (18-3n)/2=0 \quad \therefore n=6\\~\\ constant = \displaystyle \binom{9}{6}\;(2z)^{(9-6)}(\frac{-1}{\sqrt z})^6\\~\\ constant = \displaystyle \binom{9}{6}\;(2z)^3 (\frac{-1}{\sqrt z})^6\\~\\ constant = \displaystyle \binom{9}{6}\;*\;8\\~\\$$

Oct 10, 2022