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# idk

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

Jun 9, 2020

#1
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Each term is of the form
$$\binom{9}{k} (2z)^{9-k}\Big(-\frac{1}{\sqrt{z}}\Big)^k = \binom{9}{k}(-1)^k 2^{9 - k} \frac{z^{9-k}}{z^{k/2}}.$$

Therefore, the constant term occurs when 9-k=k/2 , or k=6. Putting k=6 in our expression above, the constant term is

$$\binom{9}{6} (-1)^6 2^{9 - 6} = \boxed{672}.$$

Jun 9, 2020