Page 1 of 2 12 LastLast
Results 1 to 15 of 17

Math Help - Binomial Expansion

  1. #1
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1

    Binomial Expansion

    The question is to find the first three terms of (1-9x+x^2)^5. It has a hint provided which says: 'Write it as (1-9x(1-x/9))^5 and use the binomial theorem.'

    The problem is when using the binomial equation, how should I split it up? What should be my 'x' and what should be my constant?

    (Sorry, Latex doesn't work at the moment, so it might appear a bit messy)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2008
    Posts
    792
    You have to first write it as two terms, then you can apply the following:
    Code:
    n
    Sigma (n) x^(n-k)y^k
    k=0   (k)
    Btw, by (n)(k) i meant nCk....I miss LaTeX :-(

    EDIT: I was kind of vague. n represents the number of terms (like (x+y)^4, n = 4), so if you wanted the third term, you should do this:
    Code:
    (n) x^(n-2)y^2
    (2)
    EDIT2: Example: if i wanted to find the fifth term in the expansion of (x+y)^10, then:
    - Identify n = 10
    - I need fifth term, so k = 5 - 1 = 4.
    - Evaluate :-)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1
    Quote Originally Posted by Chop Suey View Post
    You have to first write it as two terms, then you can apply the following:
    Code:
    n
    Sigma (n) x^(n-k)y^k
    k=0   (k)
    Btw, by (n)(k) i meant nCk....I miss LaTeX :-(

    EDIT: I was kind of vague. n represents the number of terms (like (x+y)^4, n = 4), so if you wanted the third term, you should do this:
    Code:
    (n) x^(n-2)y^2
    (2)
    EDIT2: Example: if i wanted to find the fifth term in the expansion of (x+y)^10, then:
    - Identify n = 10
    - I need fifth term, so k = 5 - 1 = 4.
    - Evaluate :-)
    I know the formula but for:

    (1-9x(1-x/9))^5

    Should I assume x=1 and y=-9x(1-x/9) and apply the

    (x+y)^5 binomial rule?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,617
    Thanks
    1581
    Awards
    1
    I have a complaint with the wording of this problem: “Who is to say what are the first three terms in the expansion?” You text material may define it. But I can assure you that there is no general agreement on the answer.
    I could argue that the answer is: 1+(-9x)^5+(x^2)^5.
    The general term is: [5!/(a!*b!*c!)][(1)^a][(-9x)^b][(x^2)^c] where a+b+c=5.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1
    Quote Originally Posted by Plato View Post
    I have a complaint with the wording of this problem: “Who is to say what are the first three terms in the expansion?” You text material may define it. But I can assure you that there is no general agreement on the answer.
    I could argue that the answer is: 1+(-9x)^5+(x^2)^5.
    The general term is: [5!/(a!*b!*c!)][(1)^a][(-9x)^b][(x^2)^c] where a+b+c=5.
    The first three terms means write upto and including x^2.

    As you will get:

    [constant]+[coefficient]x+[coefficient]x^2 (The first 3 terms)

    (According to textbook)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Jun 2008
    Posts
    792
    Quote Originally Posted by Air View Post
    I know the formula but for:

    (1-9x(1-x/9))^5

    Should I assume x=1 and y=-9x(1-x/9) and apply the

    (x+y)^5 binomial rule?
    Yep :-)

    (1-9x+x^2)^5 = (1-(9x-x^2))^5

    Let u = 9x-x^2

    (1-u)^5

    Sorry for delay, I dozed off, hehe.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by Air View Post
    I know the formula but for:

    (1-9x(1-x/9))^5

    Should I assume x=1 and y=-9x(1-x/9) and apply the

    (x+y)^5 binomial rule?
    Yes

    Specifically :

    (1+x)^n=\sum_{k=0}^{\infty} C_n^k x^k
    Last edited by Moo; July 28th 2008 at 10:35 AM. Reason: latex is back
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,617
    Thanks
    1581
    Awards
    1
    Quote Originally Posted by Air View Post
    The first three terms means write upto and including x^2.
    [constant]+[coefficient]x+[coefficient]x^2 (The first 3 terms)(According to textbook)
    Why did you not tell us that to begin with?
    1+{(5)(-9x)}+{[5!/(3!2!)](-9x)^2]+5x^2}
    1-45x+815x^2
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1
    I managed to do the question and others similar to the one posted earlier but this is another question that I have come upon and I am struggling on it:

    Find the coefficient of x^5 of (1+x+x^2)^4

    Can someone explain the method. Thanks in advance.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,617
    Thanks
    1581
    Awards
    1
    There are only two ways to get x^5 in that expansion:
    [(1)^0][(x)^3][(x^2)^1] & [(1)^1][(x)^1][(x^2)^2]
    The coefficient of the first is [4!/(0!)(3!)(1!)].
    The coefficient of the second is [4!/(1!)(1!)(2!)].
    Add those.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member
    Joined
    Jun 2008
    Posts
    792
    You can also do it the good ol' way


    ...

    *pants* Must...rest....now...*faints*

    P.S. Probability of a mistake in expansion: Very high.

    @Plato: is that the multinomial theorem?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1
    Quote Originally Posted by Plato View Post
    There are only two ways to get x^5 in that expansion:
    [(1)^0][(x)^3][(x^2)^1] & [(1)^1][(x)^1][(x^2)^2]
    The coefficient of the first is [4!/(0!)(3!)(1!)].
    The coefficient of the second is [4!/(1!)(1!)(2!)].
    Add those.
    I got 192 as the answer, is this correct?
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by Air View Post
    I got 192 as the answer, is this correct?
    Hmm ? Did you multiply the two coefficients Plato gave or what ? You're supposed to add them since it's like having ...+*1stCoeff*x^5+..terms of expansion..+*2ndCoeff*x^5+...

    and be careful, it's 4!/[0! 3! 1!] and 4!/[1! 1! 2!]
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1
    Quote Originally Posted by Moo View Post
    Hmm ? Did you multiply the two coefficients Plato gave or what ? You're supposed to add them since it's like having ...+*1stCoeff*x^5+..terms of expansion..+*2ndCoeff*x^5+...

    and be careful, it's 4!/[0! 3! 1!] and 4!/[1! 1! 2!]
    Oh, I thought it was (4!)/(0!) * 3! * 1! which caused me to get the answer. So, now I get 16 as the answer, is that correct?
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by Air View Post
    Oh, I thought it was (4!)/(0!) * 3! * 1! which caused me to get the answer. So, now I get 16 as the answer, is that correct?
    Yes

    I'd suggest you to get a software (like maxima or maple, but maple is not free) so that you can check by yourself or even try to check for the other coefficients ^^
    Or use a graph calculator... lol
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Binomial expansion
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: March 6th 2011, 11:20 PM
  2. Binomial Expansion
    Posted in the Pre-Calculus Forum
    Replies: 11
    Last Post: July 10th 2010, 10:00 PM
  3. binomial expansion help?
    Posted in the Algebra Forum
    Replies: 2
    Last Post: November 15th 2009, 09:21 AM
  4. Binomial expansion
    Posted in the Algebra Forum
    Replies: 4
    Last Post: August 12th 2009, 06:24 AM
  5. Replies: 6
    Last Post: May 1st 2009, 11:37 AM

Search Tags


/mathhelpforum @mathhelpforum