+0

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

0
227
1
+243

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

Jun 8, 2022

#1
+579
+3

$$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$$

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

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

$$= 672$$

-Vinculum

Jun 8, 2022