AS Maths: Bionomial Expansion! Please expand!

Write down and siimplify the first four terms in the binomial expansion of (1+2x)^6.

I don't get what to do it to. What is the first four terms?

The first term (by the order written, and given they haven't stated ascending of descending powers of x) is 1^6 = 1, so
(6C0)*1 + (6C1)*2x + (6C2)*4x² + (6C3)*8x³ +... which is
1 + 12x + 60x² + 160x³ +...

ok here is what i do for this stuff

write down pascals
(if you dont know it look it up , so)
go to the seventh row down, this is because 1 at the top = 0

ok then write down the numbers leaving a space inbetween (only four since it only asks for four)

Then you are going to do this
from left to right
(1) 1^6 X 2x^0 (6) 1^5 X 2x^1 (15) 1^5 X 2x^2 (you continue this till you have 4 terms)
((z) = pascals number here)
then what you do is multiply it out
so you would get

(1)1X1 = 1
(6)1 X 2x = 12x
(15)1 X 4x^2 = 60x^2 (you have to square the 2 here to get 4)
so you get 1+12x+60x^2 (ill let you finish it off for practice) you need 1 more term

I hoped i helped, if not add more details and ill see what i can do

1 (1)^6 | (2x)^0 = 1
6 (1)^5 | (2x)^1 = 12x
15 (1)^4 | (2x)^2 = 60x^2
20 (1)^3 | (2x)^3 = 160x^3
15 (1)^2 | (2x)^4 = 240x^4
6 (1)^1 | (2x)^5 = 190x^5
1 (1)^0 | (2x)^6 = 64x^6

so (1 + 2x)^6 = 1 + 12x + 60x^2 + 160x^3 + 240x^4 + 190x^5 + 64x^6

as you asked for the first four terms = 1 + 12x + 60x^2 + 160x^3

Hope i helped you out, this method i have used is the easiest to do.

(6C0 * 1^6 * (2x)^0) + (6C1 * 1^5 * (2x)^1) + (6C2 * 1^4 * (2x)^2) + (6C3 * 1^3 * (2x)^3) + ... =
(1 * 1 * 1) + (6 * 1 * 2x) + ((6! / (4!2!)) * 1 * 4x^2) + ((6! / (3!3!)) * 1 * 8x^3) + ... =
1 + 12x + 60x^2 + 160x^3 + ...