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# Help needed - pleasehelp!! :)

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

Dec 9, 2022

#1
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By the Binomial Theorem, the constant term is $$\dbinom{9}{3} \cdot 2^6 \cdot (-1)^3 = -5376$$

Dec 9, 2022
#2
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it said that it was wrong... thanks for helping, tho. :)

Guest Dec 9, 2022
#3
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A general term in the expansion is $$\begin{pmatrix} 9\\ n \end{pmatrix} (2z)^nz^{-(\frac{9-n}{2})}(-1)^{9-n}$$

The constant term means the z terms don't appear, which means we must have $$n-\frac{9-n}{2}=0$$  or n = 3

So the constant is given by $$\begin{pmatrix} 9\\ 3 \end{pmatrix}2^3(-1)^6=672$$

Dec 9, 2022