How do you find the binomial expansion of #(x + y)^7#?

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1 Answer
George C.
Aug 31, 2015

Use the Binomial Theorem and Pascal's triangle to find:

#(x+y)^7#

#= x^7+7x^6y+21x^5y^2+35x^4y^3+35x^3y^4+21x^2y^5+7xy^6+y^7#

Explanation:

The Binomial Theorem tells us:

#(x+y)^N = sum_(n=0)^N ((N),(n)) x^(N-n) y^n#

where #((N),(n)) = (N!)/(n! (N-n)!)#

So in our case:

#(x+y)^7 = ((7),(0))x^7 + ((7),(1))x^6y + ... + ((7),(6))xy^6 + ((7),(7))y^7#

These #((7),(n))# coefficients occur as the 8th row of Pascal's triangle (or 7th if you choose to call the first row the 0th one as some people do). Anyway, I mean the row that starts #1, 7#...

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Each number is formed by adding the two numbers above to the left and right.

The last row gives us the coefficients we need:

#(x+y)^7#

#= x^7+7x^6y+21x^5y^2+35x^4y^3+35x^3y^4+21x^2y^5+7xy^6+y^7#

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