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Math Help - binomial expansion help?

  1. #1
    Super Member Quacky's Avatar
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    binomial expansion help?

    Ok I have a few questions here and I've tried to solve them all myself, could you please check?

    2)Given that |x|<4, find in ascending powers upto and including the term x^3, the series expansion of
    a) (4-x)^1/2
    b) (1+2x)(4-x)^1/2

    3)a) Find the first four terms of the expansion in ascending powers of x of (2+3x)^-1, |x|<2/3
    b)Hence or otherwis, find the first four non-zero terms of the expansion, in ascending powers of x, of
    (1+x)(2+3x)^-1, |x|<2/3

    4)f (x)=(1+3x)^-1, |x|<1/3
    a)Expand f(x) in ascending powers of x up to and including the term in x^3
    b) Hence show for small x that
    (1+x)(1+3x)^-1 = 1-2x+6x^2-18x^3

    Thanks in advance for any replies.
    Last edited by Quacky; November 15th 2009 at 11:36 AM. Reason: 2 questions have been solved.
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  2. #2
    Junior Member
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    Quote Originally Posted by Quacky View Post
    I don't know if this is the correct thread as the titles can be so confusing, so I apologise If I've got this wrong.

    Ok I have a few questions here and I've tried to solve them all myself unsuccessfully. I know it would take hours to finish them all, so could you at least provide me with a helpful tutorial as I'll have to go to bed now and these are due first thing tomorrow :

    Finally,
    5) When (1+ax)^n is expanded as a series in asceding powers of x, the co-efficients of x and x^2 are -6 and 27 respectively. Find the values of a and n, the coefficient of x^3 and state the expressions for which the expansion is valid.

    Thanks in advance for any replies.
    i'll try the last one...
    expand the equation,
    (1+ax)^n=nC0 (1)^n \left(ax\right)^0+ nC1 (1)^{(n-1)} (ax)^1+nC2 (1)^{(n-2)} (ax)^2+..... Note: nCr=\frac {n!}{(n-r)! r!}

    (the co-efficients of x are -6)
    nC1 (1)^{(n-1)} (ax)^1=6x Note: 1^n=1
    \frac {n!}{(n-1)! 1!} (1) (a) (x)=(6) (x) (cancel x)
    \frac {n (n-1)!}{(n-1)!} (a) =6 (cancel (n-1)!)
    so,  (n) (a) = 6........... (equation 1)

    (the co-efficients of x^2 and 27 )
    nC2 (1)^{(n-2)} (ax)^2=27
    \frac {n!}{(n-2)! (2!)} (1) (a^2) (x^2) =27 (x^2) (cancel x^2)
    \frac {n (n-1) (n-2)!}{(n-2)! (2) (1)} (a^2) = 27 (cancel (n-2)!)
    \frac {n (n-1)}{2} a^2 = 27
    (n^2-n)(a^2)=54........(equation 2)

    Use the substitution method to solve (equation 1) and (equation 2) and find the value of n and a....

    You can find the coefficient for x^3 easily, after u find the value of a and n....

    ....hope it help....
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  3. #3
    Super Member Quacky's Avatar
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    Thanks, I'll have another go at the other 4 questions now.
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