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Math Help - [SOLVED] binomial theorem =(

  1. #1
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    [SOLVED] binomial theorem =(

    hey, ive got a question here in which ive been pondering over for hours, hope someone could help me ^^

    If the coefficients of  x^k and x^k+1 in the expansion of  (2 + 3x)^(10) are equal, find K.

    thanks in advance =D
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  2. #2
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    Quote Originally Posted by teddybear67 View Post
    hey, ive got a question here in which ive been pondering over for hours, hope someone could help me ^^

    If the coefficients of  x^k and x^{k+1} in the expansion of  (2 + 3x)^{10} are equal, find K.

    thanks in advance =D
    T_n is the coefficient of nth term
     <br />
T_k= T_{k+1}<br />

     <br />
^nC_k 2^k 3^{n-k} = ~^nC_{k+1} 2^{k+1} 3^{n-k-1}<br />

    Put
    n= 10
    And find k
    Last edited by ADARSH; February 17th 2009 at 04:27 AM.
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  3. #3
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    yea, thats where i got lost.
    btw sorry the 10 is supposed to be 19, so its
    <br />
(2 + 3x)^{19}<br />
    i dont think it would make any difference in the working u gave me anyways. lols
    anywyas, my working is
    <br />
^{19}C_k 2^k 3^{n-k} = ^{19}C_{k+1} 2^{k+1} 3^{n-k-1}
    =
    \frac{^{19}C_k} {^{19}C_{k+1}} = 2 X 3

    \frac{^{19}C_k} {^{19}C_{k+1}} = 6

    and i got stuck here...
    Last edited by teddybear67; February 17th 2009 at 05:35 AM.
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  4. #4
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    {19 \choose 9} = {19 \choose 10}
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  5. #5
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    i dont get it...
    i know what {19 \choose 9} = {19 \choose 10} is
    buthow did u get 9 and 10?
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  6. #6
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    don mean to post .
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  7. #7
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    Binomial Coefficients

    Hello teddybear67
    Quote Originally Posted by teddybear67 View Post
    yea, thats where i got lost.
    btw sorry the 10 is supposed to be 19, so its
    <br />
(2 + 3x)^{10}<br />
    i dont think it would make any difference in the working u gave me anyways. lols
    anywyas, my working is
    <br />
^{19}C_k 2^k 3^{n-k} = ^{19}C_{k+1} 2^{k+1} 3^{n-k-1}
    =
    \frac{^{19}C_k} {^{19}C_{k+1}} = \color{red}{2 X 3}

    \frac{^{19}C_k} {^{19}C_{k+1}} = 6

    and i got stuck here...
    You're OK to start with. But it should be

    \frac{^{19}C_k} {^{19}C_{k+1}} =\frac{2}{3}

    \Rightarrow \frac{19!(k+1)!(19-[k+1])!}{k!(19-k)!19!} = \frac{2}{3}


    \Rightarrow \frac{k+1}{19-k}=\frac{2}{3}, if you cancel carefully

    \Rightarrow 3k+3 = 38 - 2k

    \Rightarrow k = 7

    Grandad
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  8. #8
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    thanks alot grandad, but i dont get this part :
    <br />
\Rightarrow \frac{19!(k+1)!(19-[k+1])!}{k!(19-k)!19!} = \frac{2}{3}<br />

    sorry, im confused over the factorial and the "choose". its the 1 part of binomial theorem in which i never get ><

    oh ya, and the answer at the back of the book in which i got the question from is 11 and not 7. Do you know why? or is the book wrong? because there are times in which the book gives wrong answers...
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  9. #9
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    Quote Originally Posted by teddybear67 View Post
    thanks alot grandad, but i dont get this part :
    <br />
\Rightarrow \frac{19!(k+1)!(19-[k+1])!}{k!(19-k)!19!} = \frac{2}{3}<br />

    sorry, im confused over the factorial and the "choose". its the 1 part of binomial theorem in which i never get ><

    oh ya, and the answer at the back of the book in which i got the question from is 11 and not 7. Do you know why? or is the book wrong? because there are times in which the book gives wrong answers...

    Use that fact that {n \choose r}=\frac{n!}{(n-r)!r!}
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  10. #10
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    Binomial Coefficients

    Hello teddybear67
    Quote Originally Posted by teddybear67 View Post
    thanks alot grandad, but i dont get this part :
    <br />
\Rightarrow \frac{19!(k+1)!(19-[k+1])!}{k!(19-k)!19!} = \frac{2}{3}<br />

    sorry, im confused over the factorial and the "choose". its the 1 part of binomial theorem in which i never get ><

    oh ya, and the answer at the back of the book in which i got the question from is 11 and not 7. Do you know why? or is the book wrong? because there are times in which the book gives wrong answers...
    Yes, it does look rather a mess, doesn't it?

    The working goes like this:

    \frac{^{19}C_k}{^{19}C_{k+1}}=\frac{2}{3} (I assume you agree that this is correct!)

    \Rightarrow \frac{19!}{k!(19-k)!}\div \frac{19!}{(k+1)!(19-[k+1]!)} =\frac{2}{3}

    \Rightarrow \frac{19!}{k!(19-k)!}\times \frac{(k+1)!(19-[k+1]!)}{19!} =\frac{2}{3}

    Now the 19! on the top and bottom cancel immediately. Then notice that the (k+1)! term on the top = (k+1) \times k!. So it cancels with the k! term on the bottom to leave (k+1) on the top.

    Similarly the (19-k)! term on the bottom is (19-k) \times (19-[k+1])!. So it cancels with the term on the top to leave (19-k) on the bottom.

    Hence \frac{k+1}{19-k}=\frac{2}{3}

    ... etc

    I have checked my answer using an Excel spreadsheet (see the attached) and it is correct! So the answer in the book is wrong (nothing new there!).

    Grandad
    Attached Files Attached Files
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  11. #11
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    lol i assumed that your
    <br />
\frac{^{19}C_k}{^{19}C_{k+1}}=\frac{2}{3}<br />
    is correct, but i double checked and its supposed to be 3/2 . sorry!!

    but yea, i understand your workings and managed to get the hang of it. thanks alot grandad!!! ^^
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  12. #12
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    Binomial Coefficients

    Hello teddybear67
    Quote Originally Posted by teddybear67 View Post
    lol i assumed that your
    <br />
\frac{^{19}C_k}{^{19}C_{k+1}}=\frac{2}{3}<br />
    is correct, but i double checked and its supposed to be 3/2 . sorry!!

    but yea, i understand your workings and managed to get the hang of it. thanks alot grandad!!! ^^
    Ah, I see where k = 11 comes from now. Of course, it depends which way round you're expanding the binomial. Your original question has the expression as (2 + 3x)^{19}. And for this the answer is k = 7, and the equal coefficients are the 7th and 8th.

    But if you expand it the other way round (3x+2)^{19}, then the equal coefficients are the (19 - 7)th and (19 - 8)th i.e. the 12 th and 11th.

    It's all so simple really.

    Grandad

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  13. #13
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    yepyeps, thanks much much again grandad !
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