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Math Help - Partial Fractions & Binomial Theorem

  1. #1
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    Partial Fractions & Binomial Theorem

    Hi there, I need some help in this problem!

    Expand (1+xy)^n in ascending powers of y as far as y^2, where n is not a positive integer and x is a positive real number. State in terms of x , the range of values of y for which the expression is valid.

    Find the values of x and n if (1 + xy)^n ~ 1+y-y^2. Ans: n=1/3, x=3

    By replacing a suitable value for y, find an approximate value of 5^(1/3) Ans: 1 21/29
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  2. #2
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    Quote Originally Posted by puggie View Post
    Hi there, I need some help in this problem!

    Expand (1+xy)^n in ascending powers of y as far as y^2, where n is not a positive integer and x is a positive real number.

    State in terms of x , the range of values of y for which the expression is valid. Mr F says: Left for you to figure out.
    [snip]
    The general term in the expansion of (A + B)^n is ^nC_r A^{n-r}B^r so:

    (1 + xy)^n = ^nC_0 1^n (xy)^0 + ^nC_1 1^{n-1} (xy)^1 + ^nC_2 1^{n-2} (xy)^2 + ....

     = 1 + nxy + \frac{n(n-1)}{2} x^2 y^2 + ....

    Quote Originally Posted by puggie View Post
    [snip]
    Find the values of x and n if (1 + xy)^n ~ 1+y-y^2. Ans: n=1/3, x=3

    [snip]
    Solve:

    nx = 1 \Rightarrow x = \frac{1}{n} .... (1)

    \frac{n(n-1)}{2} x^2 = -1 \Rightarrow n(n-1) x^2 = -2 .... (2)

    Substitute (1) into (2): \frac{n(n-1)}{n^2} = -2 \Rightarrow \frac{n-1}{n} = -2 \Rightarrow n = \frac{1}{3}.

    Quote Originally Posted by puggie View Post
    [snip]
    By replacing a suitable value for y, find an approximate value of 5^(1/3) Ans: 1 21/29
    5^{1/3} = 2 \left( \frac{5}{8} \right)^{1/3} = 2 \left( 1 - \frac{3}{8} \right)^{1/3} so let y = -1/8.

    I get 1 \frac{23}{32} ....
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