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Well Jake, yahoo.answers.com is not here for solving your homework. We are here to help you when you get stuck on a problem and you do not know how to continue. So I won't write you all these terms, but I will give you a guidance how to do so.

(a+b)^n = (n 0)a^(n)b^0 + (n 1)a^(n-1)b^1 + (n 2)a^(n-2)b^2 .... + (n n)a^(0)b^(n)

(n k) means "n choose k" and is calculated as follows:

(n k) = n! / k!(n - k)!

For example, (5 3) is

(5 3) = 5! / 3!(2)! = (5 * 4 * 3 * 2 * 1) / (3 * 2 * 1)(2 * 1) = (5 * 4) / (2 * 1) = 10

So, lets do 3.) for example

(x+3y)^4 =

= [(4 0)x^(4)3y^(0)] + [(4 1)x^(3)3y^(1)] + [(4 2)x^(2)3y^(2)] + [(4 3)x^(1)3y^(3)] + [(4 4)x^(0)3y^(4)]

= [x^4] + [4x^(3)y^1] + [6x^(2)3y^(2)] + [4x^(1)3y^(3)] + [y^(4)]

written in a fancier way

= x⁴ + 4x³y + 6x²3y² + 4x3y³ + y⁴

A useful trick here is pascal's triangle, which looks like this

.................1..................

.............1......1..............

.........1......2.....1...........

.......1...3........3...1........

.....1..4......6.....4...1......

and so on. Each terms is sum of two terms above. For example, 4 is 3 + 1, 6 is 3+3 etc. You can extend pascal's triangle to infinity, and it will tell you coefficients of each term in (a+b)^n expansion.

First row is clearly (a+b)^0 = 1,

second row is (a+b)^1 = 1a + 1b

third row is (a+b)^2 = 1a^2 + 2ab + 1b^2

fourth row is (a+b)^3 = 1a^3 + 3a^2b + 3ab^2 + 1b^3