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Math Help - rth term and general term need help with formula please

  1. #1
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    rth term and general term need help with formula please

    My problem:

    Find the coefficient of the term containing x^3 in the expansion of (1 - \sqrt{x})^8

    I have to find the general term and compare it with the required term.

    I need to apply the rth formula in the expansion of (a+b)^n to find the general term.

    I figured out that n = 8 and a = 1.

    What I don't understand is how the exponent for b which equals x =  \frac{1}{2}
    TIA...

    See image:
    Attached Thumbnails Attached Thumbnails rth term and general term need help with formula please-rthterm.gif  
    Last edited by lilrhino; August 31st 2007 at 05:09 PM.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by lilrhino View Post
    My problem:

    Find the coefficient of the term containing x^3 in the expansion of (1 - \sqrt{x})^8

    I have to find the general term and compare it with the required term.

    I need to apply the rth formula in the expansion of (a+b)^n to find the general term.

    I figured out that n = 8 and a = 1.

    What I don't understand is how the exponent for b which equals x =  \frac{1}{2}
    TIA...

    See image:
    what do you have to multiply 1/2 by to get 3? that is the power of the binomial expansion you have to look for, that way you can get the term where the power of x is 3

    do you get it?
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    Quote Originally Posted by Jhevon View Post
    what do you have to multiply 1/2 by to get 3? that is the power of the binomial expansion you have to look for, that way you can get the term where the power of x is 3

    do you get it?
     \frac{1}{2} \div 6 = 3

    So 'r' has to be calculated before you can define the exponent for 'a'? I'm still confused as to where the  \frac{1}{2} comes from in the 1st place . How is that plugged into the formula?

    I'm supposed to find the coefficient of the term a^4

    I found 'r' by 10 - r + 1 = 8. 11 - 8 = 3.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by lilrhino View Post
     \frac{1}{2} \div 6 = 3

    So 'r' has to be calculated before you can define the exponent for 'a'? I'm still confused as to where the  \frac{1}{2} comes from in the 1st place . How is that plugged into the formula?

    I'm supposed to find the coefficient of the term a^4

    I found 'r' by 10 - r + 1 = 8. 11 - 8 = 3.
    \sqrt {x} = x^{1/2}

    what happens if we let r = 7 in the last line of those formulas you have?

    by the way, was that the formula you were given? with (r - 1) ?
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    Quote Originally Posted by Jhevon View Post
    \sqrt {x} = x^{1/2}

    what happens if we let r = 7 in the last line of those formulas you have?

    by the way, was that the formula you were given? with (r - 1) ?
    To answer the 1st question:

    10 - r + 1 = 7. Then I (think) 'r' would equal 4.


    To answer your 2nd question - I believe the r-1 is the exponent for 'b' and it came from the rth term formula:
    Attached Thumbnails Attached Thumbnails rth term and general term need help with formula please-rthterm_2.gif  
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by lilrhino View Post
    To answer the 1st question:

    10 - r + 1 = 7. Then I (think) 'r' would equal 4.


    To answer your 2nd question - I believe the r-1 is the exponent for 'b' and it came from the rth term formula:
    r has to 7. why would you say 4? remember, b is a term with a 1/2 power, we have to account for that. if r = 7, then r - 1 = 6 and 6*(1/2) = 3, which is the power we are after

    i've always used (a + b)^n = \sum_{r = 0}^{n} {n \choose r}a^{n - r}b^r ........using this formula (of course we forget about the sum, we just care about one term), we plug in r = 6 and we're home
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    Quote Originally Posted by Jhevon View Post
    r has to 7. why would you say 4? remember, b is a term with a 1/2 power, we have to account for that. if r = 7, then r - 1 = 6 and 6*(1/2) = 3, which is the power we are after

    i've always used (a + b)^n = \sum_{r = 0}^{n} {n \choose r}a^{n - r}b^r ........using this formula (of course we forget about the sum, we just care about one term), we plug in r = 6 and we're home
    I reversed the numbers - that's how I got 4. I've never seen that formula before in any of my books. I'll study it and see if I can understand it any better than the one I'm currently using. Thanks for your detailed response
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by lilrhino View Post
    I reversed the numbers - that's how I got 4. I've never seen that formula before in any of my books. I'll study it and see if I can understand it any better than the one I'm currently using. Thanks for your detailed response
    it's the same formula, except you have r-1 where i have r, and our series would start at different numbers (yours would start at r = 1)
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