Results 1 to 5 of 5

Math Help - Bionomial thoerem

  1. #1
    Newbie
    Joined
    Nov 2009
    Posts
    11

    Exclamation Bionomial thoerem

    write down and simplify the expression of (1-p)^5 .Use this result to find the expansion of (1-x-x^2)^5 in ascending powers of x as far as the terms in x^3 . Find the value of x which would enable you to estimate (0.9899)^5 from this expansion
    Follow Math Help Forum on Facebook and Google+

  2. #2
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by raza9990 View Post
    write down and simplify the expression of (1-p)^5 .Use this result to find the expansion of (1-x-x^2)^5 in ascending powers of x as far as the terms in x^3 . Find the value of x which would enable you to estimate (0.9899)^5 from this expansion
    (1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + ... + \frac{n!}{n!}x^n

    In your case it will be a finite sequence.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2009
    Posts
    11
    Quote Originally Posted by e^(i*pi) View Post
    (1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + ... + \frac{n!}{n!}x^n

    In your case it will be a finite sequence.


    but the question isn't solved yet . I tried it.
    can you write up the solution steps for me .
    I'm still confused .
    Follow Math Help Forum on Facebook and Google+

  4. #4
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by raza9990 View Post
    write down and simplify the expression of (1-p)^5 .Use this result to find the expansion of (1-x-x^2)^5 in ascending powers of x as far as the terms in x^3 . Find the value of x which would enable you to estimate (0.9899)^5 from this expansion
    I'll expand the first part for you and give guidance on the others but I'm not doing your work for you in full.

    <br />
(1-p)^5 = 1 - 5p + \frac{5 \cdot 4}{2!}p^2 - \frac{5 \cdot 4 \cdot 3}{3!}p^3 + \frac{5 \cdot 4 \cdot 3 \cdot 2}{4!}p^4 - \frac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{5!}p^5

    Any further terms will have multiplication by 0 so there are no more terms. I'll leave simplifying that to you.

    What the next question is asking is to substitute (x+x^2) for p in the above expansion. You can either do this as above or substitute p = x(1+x), if you use the latter remember that the whole expression is squared.

    For the third part it is asking you to solve 1-q = 0.9899 and q = x^2+x where q is a constant.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Nov 2009
    Posts
    11
    Quote Originally Posted by e^(i*pi) View Post
    I'll expand the first part for you and give guidance on the others but I'm not doing your work for you in full.

    <br />
(1-p)^5 = 1 - 5p + \frac{5 \cdot 4}{2!}p^2 - \frac{5 \cdot 4 \cdot 3}{3!}p^3 + \frac{5 \cdot 4 \cdot 3 \cdot 2}{4!}p^4 - \frac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{5!}p^5

    Any further terms will have multiplication by 0 so there are no more terms. I'll leave simplifying that to you.

    What the next question is asking is to substitute (x+x^2) for p in the above expansion. You can either do this as above or substitute p = x(1+x), if you use the latter remember that the whole expression is squared.

    For the third part it is asking you to solve 1-q = 0.9899 and q = x^2+x where q is a constant.

    thanks ...you are a savior
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. spectral thoerem question
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: May 12th 2010, 10:36 PM
  2. Bionomial Distribution
    Posted in the Statistics Forum
    Replies: 3
    Last Post: December 17th 2009, 12:06 PM
  3. Question on the Bionomial Theorem?
    Posted in the Statistics Forum
    Replies: 1
    Last Post: May 13th 2009, 05:09 PM
  4. [SOLVED] Expanding Bionomial
    Posted in the Algebra Forum
    Replies: 2
    Last Post: February 13th 2009, 09:28 AM
  5. The bionomial theorem
    Posted in the Algebra Forum
    Replies: 2
    Last Post: February 1st 2008, 03:04 AM

Search Tags


/mathhelpforum @mathhelpforum