What is the fourth term of (2x+3y)^6(2x+3y)6?

1 Answer
MrsRudolph221
Dec 7, 2016

The fourth term would be: (6!)/((3!)(3!))(2x)^3*(3y)^36!(3!)(3!)(2x)3⋅(3y)3

Explanation:

The 6th row of Pascal's triangle helps: 1, 6, 15, 20, 15, 6, 1

These numbers can be calculated with factorials: (6!) / ((0!)(6!))=16!(0!)(6!)=1
(6!)/((1!)(5!))=66!(1!)(5!)=6, (6!)/((2!)(4!))=156!(2!)(4!)=15, (6!)/((3!)(3!))=206!(3!)(3!)=20, and the rest of the terms will repeat in descending order.

Each of the numbers are coefficients multiplied by powers of the terms inside the binomial like so:
(n!)/((n-r)!(r!))(a)^(n-r)(b)^rn!(n−r)!(r!)(a)n−r(b)r where n = degree and r = term number -1.

In this case: (6!)/((3!)(3!))(2x)^3*(3y)^36!(3!)(3!)(2x)3⋅(3y)3 or 20(8x^3)(27y^3)20(8x3)(27y3) =4320x^3y^34320x3y3

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