What is the binomial expansion of (x 2y)7? 2x7 14x6y 42x5y2 70x4y3 70x3y4 42x2y5 14xy6 2y7 x7 14x6y 42x5y2 70x4y3 70x3y4 42x2y5 14xy6 y7 x7 7x6y 21x5y2 35x4y3 35x3y4 21x2y5 7xy6 y7 x7 14x6y 84x5y2 280x4y3 560x3y4 672x2y5 448xy6 128y7.

Respuesta :

You can take x = a, 2y = b and then can apply the binomial theorem.

The expansion of given expression is given by:

Option D: [tex]x^7 + 14x^6y + 84x^5y^2 + 280x^4y^3 + 560x^3y^4 + 672x^2y^5 + 448xy^6 + 128y^7[/tex] is

What is binomial theorem?

It provides algebraic expansion of exponentiated(integer) binomial.

According to binomial theorem,

[tex](a+b)^n = \sum_{i=0}^n ^nC_i a^ib^{n-i}[/tex]

How to use binomial theorem for given expression?

Taking a = x, and b =2y, we have n = 7, thus:

[tex](x+2y)^7 = \: ^7C_0x^0(2y)^7 + \: ^7C_1x^1(2y)^6 + \: ^7C_2x^2(2y)^5 + \: ^7C_3x^3(2y)^4 + \:^7C_4x^4(2y)^3 + \:^7C_5x^5(2y)^2 + \:^7C_6x^6y^1 + \: ^7C_0x^7y^0\\\\ (x+2y)^7 = 128y^7 + 448xy^6 + 672x^2y^5 + 560x^3y^4 + 280x^4y^3 + 84x^5y^2 + 14x^6y + x^7\\\\ (x+2y)^7 = x^7 + 14x^6y + 84x^5y^2 + 280x^4y^3 + 560x^3y^4 + 672x^2y^5 + 448xy^6 + 128y^7[/tex]

Thus, Option D: [tex]x^7 + 14x^6y + 84x^5y^2 + 280x^4y^3 + 560x^3y^4 + 672x^2y^5 + 448xy^6 + 128y^7[/tex] is correct.

Learn more about binomial theorem here:

https://brainly.com/question/86555

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