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Thread: Binomial Theorem

  1. #1
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    Binomial Theorem

    I am having big problems with the second part of this question.

    a) find a positive integer k such that

    $\displaystyle
    (x^2)^{20-k}(\frac{1}{x})^k=x
    $

    I have done this using the rules of powers and get k=13

    b) Using the binomial theorem and the answer from a) determine the coefficient of x in the expansion of.

    $\displaystyle
    (2x^2-\frac{1}{x})^{20}
    $

    I don't see how the answer from part a) can help
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  2. #2
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    Quote Originally Posted by billballest
    I am having big problems with the second part of this question.
    a) find a positive integer k such that
    $\displaystyle
    (x^2)^{20-k}(\frac{1}{x})^k=x
    $
    I have done this using the rules of powers and get k=13

    b) Using the binomial theorem and the answer from a) determine the coefficient of x in the expansion of.
    $\displaystyle
    (2x^2-\frac{1}{x})^{20}
    $
    I don't see how the answer from part a) can help
    Hello,

    If you expand this binom the first summand is $\displaystyle {20\choose 0} \cdot 2^{20} \cdot (-1)^0 \cdot x^{40}$
    The power to which x is raised will decrease by 3 per summand. As you've learned in part a) you'll get a summand with a single x if you calculate the 14th summand when k = 13.

    So the coefficient of x is: $\displaystyle {20\choose 13} \cdot 2^7 \cdot (-1)^{13} \cdot x = -9922560 x$

    Greetings

    EB
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  3. #3
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    Quote Originally Posted by earboth
    Hello,

    If you expand this binom the first summand is $\displaystyle {20\choose 0} \cdot 2^{20} \cdot (-1)^0 \cdot x^{40}$
    The power to which x is raised will decrease by 3 per summand. As you've learned in part a) you'll get a summand with a single x if you calculate the 14th summand when k = 13.

    So the coefficient of x is: $\displaystyle {20\choose 13} \cdot 2^7 \cdot (-1)^{13} \cdot x = -9922560 x$

    Greetings

    EB
    Forgive my stupidity where did the $\displaystyle x^2$ go
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  4. #4
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    Quote Originally Posted by billballest
    Forgive my stupidity where did the $\displaystyle x^2$ go
    Cuz he had,
    $\displaystyle (x^2)^{20}=x^{40}$
    By one of the rule of exponents.
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  5. #5
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    $\displaystyle
    (2x^2-\frac{1}{x})^{20}
    $

    Sorry I meant the $\displaystyle x^2
    $

    In $\displaystyle 2x^2
    $
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  6. #6
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    Quote Originally Posted by billballest
    $\displaystyle
    (2x^2-\frac{1}{x})^{20}
    $
    Sorry I meant the $\displaystyle x^2$
    In $\displaystyle 2x^2$
    Hello,

    as you may know a binom is the power of a sum, which is a sum of itself:

    $\displaystyle (a+b)^n=\sum_{i=0}^{n}{n\choose i}\cdot a^{n-i} \cdot b^i$ = $\displaystyle {n\choose 0}\cdot a^n\cdot b^0+{n\choose 1}\cdot a^{n-1}\cdot b^1+{n\choose 2}\cdot a^{n-2}\cdot b^2$+...+$\displaystyle {n\choose n}\cdot a^{0}\cdot b^n$

    So with your problem the first summands are:
    $\displaystyle {20\choose 0}\cdot 2^{20} \cdot x^{2\cdot 20}\cdot (-1)^0\cdot (x^{-1})^0$ + $\displaystyle {20\choose 1}\cdot 2^{19} \cdot x^{2\cdot 19}\cdot (-1)^1\cdot (x^{-1})^1$ + $\displaystyle {20\choose 2}\cdot 2^{18} \cdot x^{2\cdot 18}\cdot (-1)^2\cdot (x^{-1})^2+...$

    As you can see the exponents of the first factor decrease in steps of 2, put together with the negative exponents of the 2nd factor the complete exponent of x decreases by 3.

    The highest exponent is 40. The decreasing can be described by:
    Exponent = 40 - 3*i. You'll get an exponent of 1 only if i = 13 (because: 40-39 = 1)

    I hope this will help a little bit further.

    Greetings

    EB
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  7. #7
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    Quote Originally Posted by earboth
    I hope this will help a little bit further.

    Greetings

    EB
    It did thanks very much for taking the time to explain
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