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Binomial Theorem Problem

  1. Mar 26, 2014 #1
    Hello,

    I have a problem regarding the binomial theorem and a number of questions about what I can and can't do.

    1. The problem statement, all variables and given/known data

    Write the binomial expansion of [itex](1 + x)^{2}(1 - 5x)^{14}[/itex] as a series of powers of [itex]x[/itex] as far as the term in [itex]x^{2}[/itex]

    2. Relevant equations



    3. The attempt at a solution

    I know how to expand each bracket separately but i'm really unsure of how to proceed with one multiplied by the other.

    Do I expand the first one, and have that as a factor of every term in the expansion of the second?

    i.e [itex](1 + x)^{2} = 1 + x^{2} + 2x[/itex]

    [itex](1 - 5x)^{14} \approx 1 - 70x - 455x^{2}[/itex]

    [itex](1 + x)^{2}(1 - 5x)^{14} \approx 1(1 + x^{2} + 2x) - 70x(1 + x^{2} + 2x) - 455x^{2}(1 + x^{2} + 2x)[/itex]

    [itex](1 + x)^{2}(1 - 5x)^{14} \approx 1 + 2x + x^{2} - 70x - 140x^{2} - 70x^{3} - 455x^{2} - 910x^{3} - 455x^{4}[/itex]

    I get the feeling that this is wrong, but I can't find any similar examples in my text book or notes. If this happens to be the correct method, have I included to high powers? The individual expansions only reach [itex]x^{2}[/itex], but when they are combined, clearly it goes higher.

    Thanks for any help you can give!
     
    Last edited: Mar 26, 2014
  2. jcsd
  3. Mar 26, 2014 #2

    vela

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    The ##x^2## term in the expansion for ##(1-5x)^{14}## should be positive. Your work is otherwise okay, but you didn't need to calculate the ##x^3## and ##x^4## terms. You just want to identify which products will result in terms of order ##x^2## or lower and keep track of those.
     
  4. Mar 26, 2014 #3
    Ah, I forgot to square the coefficient. It should be;

    [itex](1 - 5x)^{14} \approx 1 - 70x + 2275x^{2}[/itex]

    wrt the rest of your post, does that mean I should approximate the two expansions only to the 'x' terms? Or do as before and ignore the higher powers?

    Thanks for the help.
     
  5. Mar 26, 2014 #4

    vela

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    You have to keep up to at least the ##x^2## terms because they will contribute to the final result.
     
  6. Mar 26, 2014 #5
    So,

    [itex](1 + x)^{2}(1 - 5x)^{14} \approx 1(1 + 2x + x^{2}) - 70x(1 + 2x + x^{2}) + 2275x^{2}(1 + 2x + x^{2})[/itex]

    [itex](1 + x)^{2}(1 - 5x)^{14} \approx 1 + 2x + x^{2} - 70x - 140x^2 + 2275x^{2}[/itex]

    [itex](1 + x)^{2}(1 - 5x)^{14} \approx 1 - 68x + 2136x^2[/itex]

    I think this is what you meant when you said to keep track of the products that would give me [itex]x^2[/itex] and lower.
     
  7. Mar 26, 2014 #6

    vela

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    Yup, and you can streamline it a bit further:
    $$(1 + x)^{2}(1 - 5x)^{14} \approx 1(1 + 2x + x^{2}) - 70x(1 + 2x) + 2275x^{2}(1)$$
     
  8. Mar 26, 2014 #7
    Cool - Thank you
     
  9. Mar 26, 2014 #8

    PeroK

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    That's correct. Although it might be interesting to consider for what values of x that approximation is accurate!
     
  10. Mar 26, 2014 #9
    My expansion for [itex](1 + x)^{2}[/itex] was exact, but not every term of it was used when mutliplying with the second expansion.

    My expansion for [itex](1 - 5x)^{14}[/itex] holds provided that [itex] -1 < -5x < 1[/itex] so [itex] \frac{1}{5} > x > - \frac{1}{5}[/itex]

    I don't know how I combine this information.
     
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