1 Answer

Leng Weh Seng · Answer 1 · 2021-01-12T11:01:42+0000

Answer:

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

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:

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

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

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