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

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

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

    Thanks in advance for any replies.
    Last edited by Quacky; Nov 15th 2009 at 12:36 PM. Reason: 2 questions have been solved.
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  2. #2
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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 $\displaystyle (1+ax)^n$ is expanded as a series in asceding powers of x, the co-efficients of x and $\displaystyle x^2 $are -6 and 27 respectively. Find the values of a and n, the coefficient of $\displaystyle 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,
    $\displaystyle (1+ax)^n=nC0 (1)^n \left(ax\right)^0+ nC1 (1)^{(n-1)} (ax)^1+nC2 (1)^{(n-2)} (ax)^2+.....$ Note: $\displaystyle nCr=\frac {n!}{(n-r)! r!}$

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

    (the co-efficients of$\displaystyle x^2 $ and 27 )
    $\displaystyle nC2 (1)^{(n-2)} (ax)^2=27$
    $\displaystyle \frac {n!}{(n-2)! (2!)} (1) (a^2) (x^2) =27 (x^2)$ (cancel x^2)
    $\displaystyle \frac {n (n-1) (n-2)!}{(n-2)! (2) (1)} (a^2) = 27$ (cancel (n-2)!)
    $\displaystyle \frac {n (n-1)}{2} a^2 = 27$
    $\displaystyle (n^2-n)(a^2)=54$........(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 $\displaystyle x^3$ 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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