Simple Fractions Question

Discussion in 'Math' started by TheTechGod, Sep 10, 2005.

  1. TheTechGod

    Thread Starter New Member

    Jun 15, 2005
    8
    0
    I'm studying for a test which will contain questions of how to obtain binomial coefficients, I understand the idea and can do it when the equation does not contain fractions but have trouble when it does contain fractions.

    How does one go from:

    (2x)^(9-k) (3/x)^k

    to...

    2^(9-k) 3^k x^(9-2k)

    ???

    I'm sure it's an easy year 8 level problem, But I have not worked with fractions for a long time and seem to have forgotten it! :(

    Thanks in advance!


    PS:
    For anyone actually interested in the binomial question, it is as follows:

    Find the coefficient of x in the expansion of (2x + 3/x)^9

    I have the answer if anyone wants it, I just don't understand the fractions part which I asked about above.
     
  2. haditya

    Senior Member

    Jan 19, 2004
    220
    0
    it is equivalent to finding the coeff of x^10 in the expansion of (2*x*x+3)^9

    this involves getting the 1/x term out of the expansion to the foll:
    x^-9(2*x*x+3)^9

    put x*x=y
    exp=(2*y+3)^9

    req answer= 9C5*(2^5)*(3^4)...sorry i goofed up here
     
  3. TheTechGod

    Thread Starter New Member

    Jun 15, 2005
    8
    0
    This has actually confused me even more :huh:

    The coefficient of x in my original question is 326592.

    But anyway, that is not what I'm asking about.

    The full question with the worked out answer is:

    Find the coefficient of x in the expansion of (2x + 3/x)^9

    For this expression, a general term is:

    (9 C K) (2x)^(9 - k) (3/x)^k

    This can be rewritten as

    (9 C K) 2^(9 - k) 3^k x^(9 - 2k)

    To obtain a exponent of 1, we require 9 - 2k = 1, or k = 4. Then we have:

    (9 C 4) (2^5) (3^4) = 126(32)(81) = 326592.

    Now, I understand all that... except the part where (2x)^(9 - k) (3/x)^k changed to 2^(9 - k) 3^k x^(9 - 2k).

    I understand why it was changed, but not how it was done.

    I also realise that this is a simple year 8 fractions problem, and that I should know it!

    Can anyone help?

    Thanks.
     
  4. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    I'm guessing that haditya meant to write
    As it turns out C(9,5) = C(9,4) so you both arrived at the same answer.

    Using the Binomial Series method I came up with the general espression for the set of coefficients associated with the given equation.

    C(9,K)*(2)^K*(x)^K*(3)^(9-K)*(x)^-(9-K)

    Taking the exponents of x in the above expression and summing them together and then setting them equal to 1 we get:

    K+ -(9-K) = 1
    2(K)-9 = 1
    2(K) = 10
    K = 5

    Plug in K=5 in the the general expression for the set of coefficents and I get

    C(9,5)(2^5)(x^5)(3^4)(x^-4)

    126*32*81*x = 326592x

    hgmjr
     
  5. TheTechGod

    Thread Starter New Member

    Jun 15, 2005
    8
    0
    Hi hgmjr,

    Can you explain how you got this? The way we have been taught, gives us the following:

    (9 C K) (2x)^(9 - k) (3/x)^k

    I know they are both equivalent, but the fraction confuses me. So it would help me a lot if I understood your method.

    Thank's for your help!
     
  6. haditya

    Senior Member

    Jan 19, 2004
    220
    0
    oh ok tha law of incides u mean
    1. a^m*a^n=a^(m+n)
    2. a^m/a^n=a^(m-n)
    3. (a*B)^m= a^m*b^m
    4. (a^m)^n=a^(mn)
     
  7. TheTechGod

    Thread Starter New Member

    Jun 15, 2005
    8
    0
    Hi haditya,

    I think I actually understand it now. I was digging up stuff on fractions when I should have been looking for indices.

    So...

    (2x)^(9 - k) (3/x)^k

    becomes

    2^(9 - k) 3^k

    Since the x's cancel...

    I'm still not sure why the teachers notes show it as:

    (2x)^(9 - k) (3/x)^k

    becomes

    2^(9 - k) 3^k x^(9 - 2k)

    But at least I can get the right answer now...

    Thank's all!
     
  8. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    Hi TheTechGod,

    The set of identities for manipulating exponents that haditya has posted contains the ones I used in arriving at my expression for the general case for the set of coefficients for your specific problem statement.

    1. a^m*a^n=a^(m+n)
    2. a^m/a^n=a^(m-n)
    3. (a*b )^m= a^m*b^m
    4. (a^m)^n=a^(mn)


    Starting with your expression

    (9 C K) (2x)^(9 - k) (3/x)^k

    It can be expanded using identity 3 above to get

    (9CK)*(2)^(9-K)*(x)^(9-K)*(3/x)^(K)

    It can be further expanded using identity 2 above to get

    (9CK)*(2)^(9-K)*(x)^(9-K)*(3)^(K)*(x)^(-K)

    Taking the exponents of the two x terms above.

    (9 - K)+(-K) = 1

    9-2K = 1
    -2K = -8
    K=4

    Then plugging K=4 the following equation is obtained

    (9C4)*(2)^(5)*(x)^(5)*(3)^(4)*(x)^(-4)

    Collecting the two x terms yields

    (9C4)*(2^5)*(3^4)*(x^5)*(x^-4)

    Rearranging and then combining the two x terms using identity 1 yields

    (9C4)*(2^5)*(3^4)*(x)

    As I mention in my earlier reply (9C4) evaluates to the same value as (9C5) so that choice of which convention to use between (9-K),(K) or (K),(9-K) yields the same result.

    Hope this helps.

    hgmjr
     
  9. haditya

    Senior Member

    Jan 19, 2004
    220
    0
    the x's dont cancel but add up.
    coz its (1/x)^k * x^(9-k)
    1/x=x^-1
    therefore the result becomes (x^-1)^k * x^(9-k)
    (x^-k) * x^(9-k)

    hence the result
     
  10. TheTechGod

    Thread Starter New Member

    Jun 15, 2005
    8
    0
    Hi everyone,

    Thanks for everything, I passed the test :)
     
Loading...