Question

The first four terms of the expansion of the binomial (2x³ + 3y²)^7 are shown
128x^21 + 1,344x¹8y² + 6,048x¹5y +15, 120x¹²y + ...
Complete the last four terms by dragging the missing values to the correct locations in t
expression. Not all values will be used.
¯¯¯¯¯¯¯x³y² + 20,412
x¹y7
x6y10
x6 1,344
3y
22, 680
10, 206
+
2,187¹4

Answer 1

Th last four terms of the binomial expansion are 14175x^9y^8 ,  20412x^6y^10  ,  10206x^3y^12  and 2187y^14.

The formula of binomial expansion is

              (x + y)^n = Σ (nCk) x^(n-k) y^k             (1)

We have bee given the expression as (2x^3 + 3y^2)^7

Here, x = 2x^3   ,    y = 3y^2   and  n = 7

Putting the required values in equation (1) we get

        (2x^3 + 3y^2)^7   =   7C0(2x^3)^7(3y^2)^0 +  7C1(2x^3)^6(3y^2)^1 +  7C2(2x^3)^5(3y^2)^2 +  7C3(2x^3)^4(3y^2)^3 + 7C4(2x^3)^3(3y^2)^4 +  7C5(2x^3)^2(3y^2)^5 + 7C6(2x^3)^1(3y^2)^6 +  7C7(2x^3)^0(3y^2)^7        

 

        (2x^3 + 3y^2)^7    =   128x^21  +  7*(64x^18)*3y^2   +  21*(32x^15)*(9y^4) +  35*(16x^12)*(21y^6) +  35*(5x^9)*(81y^8)  +  21*(4x^6)*(243y^10) +  7*(2x^3)*(729y^12) + 2187y^14

(2x^3 + 3y^2)^7  =   128x^21   + 1344x^18y^2  + 6048x^15y^4  + 11760x^12y^6  +  14175x^9y^8 +  20412x^6y^10  +  10206x^3y^12  +  2187y^14.

Hence the last four terms of binomial expansion are  14175x^9y^8 ,  20412x^6y^10 , 10206x^3y^12  and  2187y^14.

Learn more about binomial expansion here : https://brainly.com/question/13800206

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Answer 2

Answer:

In order the missing terms are:

$22680x^{9} y^{8}$

$20412x^{6}y^{10}$

$10206x^{3}y^{12}$

$2187y^{14}$