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Binomial theorem

  1. Jun 17, 2006 #1
    1. For each of the following, simplify so that the variable term is raised is to a single power:
    (a) State the General Term tr in the Binomial expansion of (2x^2 - 1/x)^10
    (b) Find the 7th term in the expansion
    (c) Is there an x^5 term? Find its coefficient.
    (d) Is there a constant term [independent of x] ? Find it if there is?

    this is what i have so far...

    a) tr = C(10,r) (2x^2)^(10-r)*(-1/x)^r
    = C(10,r) 2^(10-r) * x^(20-r)* (-1)^r * x^-r
    = 2^(10-r) * (-1)^r * C(10,r) * x^(20-3r)

    b) Since it is t7, r = 6

    so t7 = 2^4 * (-1)^6 * C(10,6) * x^2
    = 3360x^2

    c) Let 20-3r = 5, so r = 5.

    So t5 = 2^5 * (-1)^5 * C(10,5) * x^2
    and the coefficient is -8064

    d) Let 20-3r = 0
    r = 20/3 so there is no constant term?

    a somewhat unrelated topic, but another quesiton...

    2. [​IMG]

    how can i utilize the hints to develop a formula?
    Last edited: Jun 17, 2006
  2. jcsd
  3. Jun 17, 2006 #2

    matt grime

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    b) So, you can find the 'general' term, but not the 7th? r goes from 0 to 10 in a) so the 7th term is when r=6.

    c) that is not a coefficient; the coefficient is the number that multiplies the x^5 term (see d))

    d) if you can do c) you can do d) unless you just put r=5 because you saw x^5 and thought that automatically meant r=5. constant means x^0.
  4. Jun 17, 2006 #3
    thanks matt, i edited my post around the same time you posted! can you check what i have now? and any help for the second question?
  5. Jun 17, 2006 #4

    matt grime

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    I would prefer to use the statement that (x^2)^r(1/x)^(10-r) always gives an odd power of x, hence there is no x^0 term. I dislike intensely unnecessary proofs that rely on showing something supposed to be an integer isn't by using fractions. Call me an old fuss pot if you must. But it is completely unnecessary to do this as a 'contradiction', as are many contradiction proofs.
    Last edited: Jun 17, 2006
  6. Jun 17, 2006 #5


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    For the second question, find the sum of the first n even squares (this should be easy), and then subtract this off the sum of the first 2n squares.
  7. Mar 8, 2010 #6
    thanks for the information!!!! more power
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