Question

How do you expand the binomial `(x^2+y)^7` using the binomial theorem?

Answer 1

The answer is `=x^14+7x^12y+21x^10y^2+35x^8y^3+35x^6y^4+21x^4y^5+7x^2y^6+y^7`

Explanation:

Let's apply

`(a+b)^n=((n),(0))a^nb^0+((n),(1))a^(n-1)b+((n),(2))a^(n-3)b^2+.....+((n),(n))a^0b^n.`

Where, `((n),(p))=(n!)/((n-p)!p!)`

`((n),(1))=(n!)/((n-1)!(1!))=n`

In our case, `a=x^2` , `b=y` and `n=7`

`(x^2+y)^7=((7),(0))(x^2)^7+((7),(1))(x^2)^6y+((7),(2))(x^2)^5y^2+((7),(3))(x^2)^4y^3+((7),(4))(x^2)^3y^4+((7),(5))(x^2)^2y^5+((7),(6))(x^2)y^6+((7),(7))y^7`

`(x^2+y)^7=x^14+7x^12y+21x^10y^2+35x^8y^3+35x^6y^4+21x^4y^5+7x^2y^6+y^7`

`((7),(0))=1`

`((7),(1))=7`

`((7),(2))=21`

`((7),(3))=35`

`((7),(4))=35`

`((7),(5))=21`

`((7),(6))=7`

`((7),(7))=1`