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Math Help - Help with the Binomial Theorem

  1. #1
    Newbie
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    Smile Help with the Binomial Theorem

    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!!!!
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  2. #2
    Lord of certain Rings
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    Quote Originally Posted by statsnewbie View Post
    I am somewhat confused.

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

    <br />
(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k<br />


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

    <br />
\color{red}{(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}b^{n-k}a^k}<br />

    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<br />

    (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
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  3. #3
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    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]
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  4. #4
    Lord of certain Rings
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    Quote Originally Posted by statsnewbie View Post
    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
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  5. #5
    MHF Contributor kalagota's Avatar
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    nice proof isomorphism..

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