Help with the Binomial Theorem

**statsnewbie** · June 24th 2008, 04:07 AM

I am somewhat confused.

Some text books refer to the following as the "General Form"

math]
(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k
[/tex]

and others reverse the a^n and b^n-k

math]
(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k
[/tex]

Are they both correct and if not when does one use the one or the other.

Any help would be great!!!!

**Isomorphism** · June 24th 2008, 04:18 AM

Originally Posted by statsnewbie

I am somewhat confused.

Some text books refer to the following as the "General Form"

$ (a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k $

and others reverse the a^n and b^n-k

$ \color{red}{(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}b^{n-k}a^k} $

Are they both correct and if not when does one use the one or the other.

Any help would be great!!!!

The way you have written it both are the same. So I have changed it. See the above part marked in red.

Yes both are correct. That happens because $\binom{n}{k} = \binom{n}{n - k}$ .

Lets use it and prove that:

$(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k =\sum_{k=0}^{n}\binom{n}{k}b^{n-k}a^k $

$(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k$

$(a+b)^n=\sum_{k=0}^{n}\binom{n}{n - k}a^{n-k}b^k$

Now put t = n - k, the k varies from 0 to n, t varies from n to 0(thats the same as 0 to n).

$(a+b)^n=\sum_{t=0}^{n}\binom{n}{t}a^{t}b^{n-t}$

But het t is just a variable we introduced to count the powers of a, its a dummy variable. Lets replace it with k.

$(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^{k}b^{n-k}$

Hence proved

**statsnewbie** · June 24th 2008, 04:40 AM

Thanks v much.

If we can then continue, the problem I was trying to solve was the following

Find the term that does not contain x
(8x+1/2x)^8
and what I worked out was that of course it's where x=0 and n=8

and then to solve for what r is I said was

(8x)^n-r and (1/2x)^n

and then found that 8-2r=0 and r=4, both terms evaluate to 1 and it's term that has n chose r = 8! /4!.4!

Hope that it makes sense, not sure how to use the math tags properly[/SIZE]

[/SIZE]

[/SIZE][/FONT][/SIZE][/FONT]

**Isomorphism** · June 24th 2008, 05:36 AM

Originally Posted by statsnewbie

Thanks v much.

If we can then continue, the problem I was trying to solve was the following

Find the term that does not contain x
(8x+1/2x)^8
and what I worked out was that of course it's where x=0 and n=8

and then to solve for what r is I said was

(8x)^n-r and (1/2x)^n

and then found that 8-2r=0 and r=4, both terms evaluate to 1 and it's term that has n chose r = 8! /4!.4!

Hope that it makes sense, not sure how to use the math tags properly[/size]

[/size]

[/size][/font][/size][/font]

Yes you are nearly right.

The coefficient is $\binom{8}{r}(8x)^{n-r}\left(\frac1{2x}\right)^r = \binom{8}{4}(8x)^{4}\left(\frac1{2x}\right)^4 = \binom{8}{4} \frac{8^4}{2^4} = \binom{8}{4} 4^4$

**kalagota** · June 24th 2008, 05:40 AM

nice proof isomorphism..

the simplest reason one could say is that addition in R is commutative.. Ü

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