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

 Jun 8, 2022
 #1
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\(\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i\)

 

\(\sum _{i=0}^9\binom{9}{i}\left(2z\right)^{\left(9-i\right)}\left(-\frac{1}{\sqrt{z}}\right)^i\)

 

\(512z^9-2304z^{\frac{15}{2}}+4608z^6-\frac{5376z^6}{\left(\sqrt{z}\right)^3}+4032z^3-\frac{2016z^4}{\left(\sqrt{z}\right)^5}+672-\frac{144z^2}{\left(\sqrt{z}\right)^7}+\frac{18}{z^3}-\frac{1}{\left(\sqrt{z}\right)^9}\)

 

\(= 672\)

 

-Vinculum

 

smileysmileysmiley

 Jun 8, 2022

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