- #1

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**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

- Thread starter roger
- Start date

- #1

- 319

- 0

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

- 1,197

- 1

check out this site:

http://www.mathsdirect.co.uk/pure/purtutbinbas.htm

http://www.mathsdirect.co.uk/pure/purtutbinbas.htm

- #3

- #4

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[tex]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!}+...[/tex]

So in your case,

[tex]f(x) = (a+b)^x[/tex]

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

- #5

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when used in binomial theorem ?

what does it mean n choose x ?

roger

- #6

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[tex] ^n C _r [/tex]

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

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Lets just expand some simple brackets first:

[tex](a+b)^0 = 1[/tex]

[tex](a+b)^1 = a+b[/tex]

[tex](a+b)^2 = (a+b)(a+b) = a^2+ab+ba+b^2 = a^2+2ab+b^2[/tex]

[tex](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[/tex]

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 [tex](a+b)^1^5[/tex]or something in that nature.

Also it is possible to see that there is a pattern to the coefficents (the numbers before the [tex]ab[/tex] or the [tex]ab^2[/tex]). 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 [tex](a+b)^2[/tex] and work it out using the binomial theorem.

The first part is [tex]a^2[/tex]. We know this without any working out that occurs on paper but using the binomial it would be:

[tex](a)^2[/tex] which gives [tex]a^2[/tex].

The [tex]2ab[/tex] part is denoted by [tex]2(a^1b^1)[/tex] which gives [tex]2ab[/tex].

The last part of this is similar to the first but using b instead of a.

The pattern is: [tex](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[/tex]

As you can see, the indices add up to the original indice.

A harder one just to show what the binomial does:

[tex](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)[/tex]

[tex]= 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)[/tex]

[tex]= a^6 + 6a^5b + 15a^4b^2 + 20a^3b^3 + 15a^2b^4 + 6ab^5 + b^6[/tex]

To finish the reason that I put the [tex]b[/tex] term in brackets without the indice is because if you get [tex](a-b)^n[/tex] then the part of the equation could be negative (e.g. [tex]20(a^3(-b)^3) = -20a^3b^3[/tex]).

I hope this helps.

The Bob (2004 ©)

[tex](a+b)^0 = 1[/tex]

[tex](a+b)^1 = a+b[/tex]

[tex](a+b)^2 = (a+b)(a+b) = a^2+ab+ba+b^2 = a^2+2ab+b^2[/tex]

[tex](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[/tex]

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 [tex](a+b)^1^5[/tex]or something in that nature.

Also it is possible to see that there is a pattern to the coefficents (the numbers before the [tex]ab[/tex] or the [tex]ab^2[/tex]). 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 [tex](a+b)^2[/tex] and work it out using the binomial theorem.

The first part is [tex]a^2[/tex]. We know this without any working out that occurs on paper but using the binomial it would be:

[tex](a)^2[/tex] which gives [tex]a^2[/tex].

The [tex]2ab[/tex] part is denoted by [tex]2(a^1b^1)[/tex] which gives [tex]2ab[/tex].

The last part of this is similar to the first but using b instead of a.

The pattern is: [tex](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[/tex]

As you can see, the indices add up to the original indice.

A harder one just to show what the binomial does:

[tex](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)[/tex]

[tex]= 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)[/tex]

[tex]= a^6 + 6a^5b + 15a^4b^2 + 20a^3b^3 + 15a^2b^4 + 6ab^5 + b^6[/tex]

To finish the reason that I put the [tex]b[/tex] term in brackets without the indice is because if you get [tex](a-b)^n[/tex] then the part of the equation could be negative (e.g. [tex]20(a^3(-b)^3) = -20a^3b^3[/tex]).

I hope this helps.

The Bob (2004 ©)

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