Lets just expand some simple brackets first:
(a+b)^0 = 1
(a+b)^1 = a+b
(a+b)^2 = (a+b)(a+b) = a^2+ab+ba+b^2 = a^2+2ab+b^2
(a+b)^3 = (a+b)(a+b)(a+b) = (a+b)(a^2+2ab+b^2) = <br />
<br />
a^3+2a^2b+ab^2+a^2b+2ab^2+b^3 = a^3+3a^2b+3ab^2+b^3
As you can see I have expanded the brackets the normal (well normal to me) way to do so. The problem that the binomial solves is when you have (a+b)^1^5or something in that nature.
Also it is possible to see that there is a pattern to the coefficents (the numbers before the ab or the ab^2). The pattern, if wirtten out, is a pascal triangle. The top row has a single 1 then row two as two 1's (either side of the original one) and so on. This like will show you what I mean and then look back at the brackets above. The coefficents match (
http://mathworld.wolfram.com/PascalsTriangle.html).
Lets take the (a+b)^2 and work it out using the binomial theorem.
The first part is a^2. We know this without any working out that occurs on paper but using the binomial it would be:
(a)^2 which gives a^2.
The 2ab part is denoted by 2(a^1b^1) which gives 2ab.
The last part of this is similar to the first but using b instead of a.
The pattern is: (a+b)^2 = ^2 C _0(a^2(b)^0) + ^2 C _1(a^1(b)^1) + ^2 C _2(a^0(b)^2) = 1(a^2(1)) + 2(a(b)) + 1(1(b^2)) = a^2+2ab+b^2
As you can see, the indices add up to the original indice.
A harder one just to show what the binomial does:
(a+b)^6 = ^6 C _0(a^6(b)^0) + ^6 C _1(a^5(b)^1) + ^6 C _2(a^4(b)^2) + ^6 C _3(a^3(b)^3) + ^6 C _4(a^2(b)^4) + ^6 C _5(a^1(b)^5) + ^6 C _6(a^0(b)^6)
= 1(a^6(1)) + 6(a^5(b)) + 15(a^4(b)^2) + 20(a^3(b)^3) + 15(a^2(b)^4) + 6(a^1(b)^5) + 1(a^0(b)^6)
= a^6 + 6a^5b + 15a^4b^2 + 20a^3b^3 + 15a^2b^4 + 6ab^5 + b^6
To finish the reason that I put the b term in brackets without the indice is because if you get (a-b)^n then the part of the equation could be negative (e.g. 20(a^3(-b)^3) = -20a^3b^3).
I hope this helps.
The Bob (2004 ©)
