Find the constant term in the expansion of \((2z - \frac{1}{\sqrt{z}}\Big)^9.\)
By the Binomial Theorem, the constant term is \(\dbinom{9}{3} \cdot 2^6 \cdot (-1)^3 = -5376\)
+33661 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 \(\begin{pmatrix} 9\\ 3 \end{pmatrix}2^3(-1)^6=672\)
