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    I've been going through the 2015 C4 maths paper and cannot figure out how you do the second part to this question:
    Find the binomial expansion of up to and including the term in . [3 marks]

    I think the answer to this is: 1/4 - x/16 +( 5x^2)/256

    (ii) Use your expansion from part (b)(i) to find an estimate for , giving your answer to four decimal places. [2 marks]

    Any help on (ii) would be really appreciated!
    Thank you.
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    (Original post by EmJ15)
    I've been going through the 2015 C4 maths paper and cannot figure out how you do the second part to this question:
    Find the binomial expansion of up to and including the term in . [3 marks]

    I think the answer to this is: 1/4 - x/16 +( 5x^2)/256

    (ii) Use your expansion from part (b)(i) to find an estimate for , giving your answer to four decimal places. [2 marks]

    Any help on (ii) would be really appreciated!
    Thank you.
    what you do is use trial and improvement and try and make your become , by playing around with the first equation once that x value gives you the second equation what you do is sub in that value of x into your expansion and find the value
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    So the expansion is valid for  x<8/3 . What happens if you let x=1/3.
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    (Original post by EmJ15)
    I've been going through the 2015 C4 maths paper and cannot figure out how you do the second part to this question:
    Find the binomial expansion of up to and including the term in . [3 marks]

    I think the answer to this is: 1/4 - x/16 +( 5x^2)/256
    This bit is correct.

    (ii) Use your expansion from part (b)(i) to find an estimate for , giving your answer to four decimal places. [2 marks]

    Any help on (ii) would be really appreciated!
    Thank you.
    If you notice that:

    \displaystyle

\begin{equation*}\left(\frac{1}{  81}\right)^{1/3} = \left(\frac{1}{9^2}\right)^{1/3} = (9^{-2})^{1/3} = 9^{-2/3}\end{equation*}

    Then it is fairly obvious that x = 1/3 gives you (8 + 1)^{-2/3} and you can plug this into your binomial expansion to get an approximation.
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    (Original post by Zacken)
    This bit is correct.



    If you notice that:

    \displaystyle

\begin{equation*}\left(\frac{1}{  81}\right)^{1/3} = \left(\frac{1}{9^2}\right)^{1/3} = (9^{-2})^{1/3} = 9^{-2/3}\end{equation*}

    Then it is fairly obvious that x = 1/3 gives you (8 + 1)^{-2/3} and you can plug this into your binomial expansion to get an approximation.
    Thank you! That makes more sense now.
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    (Original post by EmJ15)
    Thank you! That makes more sense now.
    You're welcome!
 
 
 
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