Question

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

Answer 1

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`