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

1 Answer
Dec 7, 2016

The fourth term would be: #(6!)/((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!))=1#
#(6!)/((1!)(5!))=6#, #(6!)/((2!)(4!))=15#, #(6!)/((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)^r# where n = degree and r = term number -1.

In this case: #(6!)/((3!)(3!))(2x)^3*(3y)^3# or #20(8x^3)(27y^3)# =#4320x^3y^3#