# How do you find the fourth term of (2x+3y)^9?

$217728 {x}^{6} {y}^{3}$
Using binomial expansion, we know that the fourth term of ${\left(2 x + 3 y\right)}^{9} = \left({\text{^9"C}}_{4}\right) {\left(2 x\right)}^{6} {\left(3 y\right)}^{3} = \left(216\right) \left(64 {x}^{6}\right) \left(27 {y}^{3}\right) = 217728 {x}^{6} {y}^{3}$