Question

Find the middle term in the expansion ${(\frac{x}{2}-2y)}^6$ using pascal's triangle.

Answer 1

Term Multiplier/Pascal's Triangle Expansion Based on Exponent

${(a+b)}^0=1$

${(a+b)}^1=1-1$

${(a+b)}^2=1-2-1$

${(a+b)}^3=1-3-3-1$

${(a+b)}^4=1-4-6-4-1$

${(a+b)}^5=1-5-10-10-5-1$

${(a+b)}^6=1-6-15-20-15-6-1$

Thus, expansion ${(a+b)}^6=1a^6{b}^0+6a^5{b}^1+15a^4{b}^2+20a^3{b}^3+15a^2{b}^4+6a^1{b}^5+1a^0{b}^6$

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

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