Answer:
[tex]T_{3}=27{x}^{2}y^2[/tex]
Step-by-step explanation:
The given binomial expression is:
[tex]( {x}^{2} + 3y)^{3} [/tex]
When we compare to:
[tex] {(a +b)}^{n} [/tex]
We have
[tex]a = {x}^{2} [/tex]
[tex]b = 3y \\ n = 3[/tex]
The nth term is given by;
[tex]T_{r+1}=^nC_ra^{n-r}b^r[/tex]
To find the 3rd term, we put:
[tex]r + 1 = 3 \\ r = 2[/tex]
We substitute into the formula to get:
[tex]T_{3}=^3C_2( {x}^{2} )^{3-2}(3y)^2[/tex]
We simply:
[tex]T_{3}=3( {x}^{2} )^{1} \times 9y^2[/tex]
[tex]T_{3}=27{x}^{2}y^2[/tex]
