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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

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