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

1 Answer
Narad T.
Nov 23, 2016

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#

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