What is the middle term in the expansion of (x/2-2y)^6?

Mar 24, 2015

Since this was asked under "Pascal's Triangle and Binomial Expansion" let's use that

Term Multiplier/Pascal's Triangle Expansion Based on Exponent
${\left(a + b\right)}^{0} : 1$
${\left(a + b\right)}^{1} : 1 - 1$
${\left(a + b\right)}^{2} : 1 - 2 - 1$
${\left(a + b\right)}^{3} : 1 - 3 - 3 - 1$
${\left(a + b\right)}^{4} : 1 - 4 - 6 - 4 - 1$
${\left(a + b\right)}^{5} : 1 - 5 - 10 - 10 - 5 - 1$
${\left(a + b\right)}^{\textcolor{red}{6}} : 1 - 6 - 15 - \textcolor{red}{20} - 15 - 6 - 1$
$\textcolor{w h i t e}{\text{XXXX}}$That is, the expansion of
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$${\left(a + b\right)}^{\textcolor{red}{6}}$
$\textcolor{w h i t e}{\text{XXXX}}$is
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$1 {a}^{6} {b}^{0} + 6 {a}^{5} {b}^{1} + 15 {a}^{4} {b}^{2} + \textcolor{red}{20 {a}^{3} {b}^{3}} + 15 {a}^{2} {b}^{4} + 6 {b}^{1} {b}^{5} + 1 {a}^{0} {b}^{6}$

with $a = \frac{x}{2}$ and $b = - 2 y$
the middle term of ${\left(\frac{x}{2} - 2 y\right)}^{6}$ is

$\textcolor{red}{20} {\left(\frac{x}{2}\right)}^{3} {\left(- 2 y\right)}^{3}$

$= - 20 {x}^{3} {y}^{3}$