Question

HELP!!Find the third term of (x^2+3y)^3

Answer 1

Answer:

$T_{3}=27{x}^{2}y^2$

Step-by-step explanation:

The given binomial expression is:

$( {x}^{2} + 3y)^{3}$

When we compare to:

${(a +b)}^{n}$

We have

$a = {x}^{2}$

$b = 3y \\ n = 3$

The nth term is given by;

$T_{r+1}=^nC_ra^{n-r}b^r$

To find the 3rd term, we put:

$r + 1 = 3 \\ r = 2$

We substitute into the formula to get:

$T_{3}=^3C_2( {x}^{2} )^{3-2}(3y)^2$

We simply:

$T_{3}=3( {x}^{2} )^{1} \times 9y^2$

$T_{3}=27{x}^{2}y^2$