# Binomial Theorem

1. Dec 26, 2004

### roger

Binomial Theorem......

Hi

I need to know about binomial theorem....

eg. how to in general expand (a+b)^n

I dont understand the combinations / permutations....?

thanks

Roger

2. Dec 26, 2004

### UrbanXrisis

3. Dec 26, 2004

### Sirus

4. Dec 26, 2004

### Hyperreality

Another expansion can be obtained using the Taylor Function.

$$f_n(x)=f(0)+f'(0)x+\frac{f''(0)x^2}{2!}+\frac{f^(3)(0)x^3}{3!}+...+\frac{f^(n)(0)x^n}{n!}+...$$

$$f(x) = (a+b)^x$$

P.S. If I remember correctly, this only applies for real numbers.

5. Dec 27, 2004

### roger

what i dont understand is the combinations and permutations........
when used in binomial theorem ?

what does it mean n choose x ?

roger

6. Dec 27, 2004

### misogynisticfeminist

the thing I am used to is

$$^n C _r$$

i just use the one in my calculator, yours should have too. in fact, my teacher taught us the binomial theorem before permuations and combinations !

7. Dec 27, 2004

### The Bob

Lets just expand some simple brackets first:
$$(a+b)^0 = 1$$
$$(a+b)^1 = a+b$$
$$(a+b)^2 = (a+b)(a+b) = a^2+ab+ba+b^2 = a^2+2ab+b^2$$
$$(a+b)^3 = (a+b)(a+b)(a+b) = (a+b)(a^2+2ab+b^2) = a^3+2a^2b+ab^2+a^2b+2ab^2+b^3 = a^3+3a^2b+3ab^2+b^3$$
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 $$(a+b)^1^5$$or 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.