Binomial expansion - please help (AS Level)

Hi, I am trying to do a maths paper, and I'm totally stuck on this question. I'm so angry at myself because I know it should be really easy, but I just can't remember how to do it. I have the Answer Booklet, with the answer and the steps to it, but it doesn;t explain it so I don't understand. please can you do a step by step answer, and explain why you did each step, I'll be really grateful, thank you!

a) Find the first 3 terms, in ascending powers of x, of the binomial expansion of (1 + px)^9 where p is a constant.

b) The first 3 terms are 1, 36x and qx^2, where q is a constant. Find the value of p and q.

a) The expansion of (1+x)^n is given as 1 + nx + (n(n-1)x^2)/2 +...
Therefore (1+px)^9 = 1+9(px)+(9*8*((px)^2))/2+...
The first term is :1
Second: 9px
Third: 36(p^2)(x^2)

b) Comparing the second terms together, p can be found:
9px = 36x
p = 4

Therefore the third term in the expansion, now that we found p is (36)(16)(x^2) = 576x^2
Therefore comparing it with qx^2 we get that q = 576

In the following C(N,k) stands for N!/(k! * (N-k)!)); and is read "N-choose-k".
In general
(1 + y)^N = sum(i = 0 to N) of C(N,i) * y^i

So you first three terms are: 1 + 9(px) + (9*8)/(2*1) (px)^2 + (9*8*7)/(3*2*1) (px)^3 + ...
Now you can do the rest

let's use x and y, and when you re done, replace x=1, y= px)
for
(x+y)^9=
step one
write xy terms, without the coefficients, denoted as just empty()
(x+y)^9=()x^9(y^0)+()x^8(y^1)+()x^7(y^…
+()x^6(y^3)+()x^5(y^4)+()x^4(y^5)
+()x^3(y^6)+()x^2(y^7)+()x^1(y^8)+()x^…
recal that a^0=1
a^1=a
that is how they come up with
(x+y)^9=()x^9+()x^8(y)+()x^7(y^2)
+()x^6(y^3)+()x^5(y^4)+()x^4(y^5)
+()x^3(y^6)+()x^2(y^7)+()x(y^8)+()(y^9…

for the coefficients
the 1st term C is always=1
this C can be found from nCk formula
nCk= n!/(n-k)!k!
1st term k=0
2nd term k=1
3rd term k=2...etc
so for 2nd term
nCk=9C1=9!/(9-1)!(1!)
for factorial
0!=1
1!=1
so 9C1=[1*2*3*4*5*6*7*8*9/(1*2*3*4*5*6*7*8)…
cancel out what you can
9C1=9/1=9
so you can go to any term and find its coefficient
if you want quick approach, and allow to use the net, google for pascal's triangle, they provide you with coefficient table of binomial expansion for some n
after you write down all the info needed, replace x=1, and y=px, simplify it as you can