Binomial expansion - please help (AS Level)?

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

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^2)

+()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^0(y^9)

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)(1)

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

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

the 1st 3 words of a binomial growth (q+p)^n are nC0*q^(n-0)*p^0 , nC1*q^(n-a million)*p^a million and nC2*q^(n-2)*p^2 in this question q = a million, p = px and n = 9 2d term is 9C1*a million^8*(px)^a million = 36x 9*a million*px = 36x 9px = 36x p = 36x/9x p = 4 0.33 term is 9C2*a million^7*(px)^2 = qx^2 36*a million*p^2x^2 = qx^2 36p^2x^2 = qx^2 36p^2 = q q = 36*4^2 = 36*sixteen = 576