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Math Help - 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

    <br />
(x^2)^{20-k}(\frac{1}{x})^k=x<br />

    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.

    <br />
(2x^2-\frac{1}{x})^{20}<br />

    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
    <br />
(x^2)^{20-k}(\frac{1}{x})^k=x<br />
    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.
    <br />
(2x^2-\frac{1}{x})^{20}<br />
    I don't see how the answer from part a) can help
    Hello,

    If you expand this binom the first summand is {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: {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 {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: {20\choose 13} \cdot 2^7 \cdot (-1)^{13} \cdot x = -9922560 x

    Greetings

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

    Sorry I meant the x^2<br />

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

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

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

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