The greatest term in the expansion of $(2x + 3y)^{11}$ when x = 9 and y = 4 is :

Given:- $(2 x+3 y)^{\prime \prime}$ at $x=9, y=4$

The Greatest term will be

$\begin{array}{l} =\frac{(n+1) \times|y|}{|x|+|y|} \\ =\frac{12 \times 4}{9+4}=\frac{48}{13}=3.69 \end{array}$

Hence, Greatest term is $4^{\text {th }}$ term

$\begin{aligned} T_{4}=T_{3+1} & ={ }^{H} C_{3}(2 x)^{8}(3 y)^{3} \\ & =\frac{11 \times 10 \times 9 \times 8 !}{3 \times 2 \times 1 \times 8 !}(2 \times 9)^{8}(3 \times 4)^{3} \\ T_{4} & =165(18)^{8}(12)^{3} \end{aligned}$