Results 1 to 6 of 6

Math Help - Sigma notation & Pascal's Triangle/Binomial Theorem.

  1. #1
    Junior Member
    Joined
    Jan 2007
    Posts
    26

    Question Sigma notation & Pascal's Triangle/Binomial Theorem.

    I have no idea how to do these:

    A) "Find the sum of the arithmetic series"




    B) "Expand (x+2)^4"

    So far, all I could manage to get is 1x^4(2)^0+4/1x^3(3). Is there an easier way to do this question without it becoming so messy/confusing?

    Thanks in advance!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie Logarithm Freak's Avatar
    Joined
    May 2007
    From
    Houston, Texas
    Posts
    3
    Sn= (N/2)(A1 + An)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,845
    Thanks
    320
    Awards
    1
    Quote Originally Posted by Mulya66 View Post
    B) "Expand (x+2)^4"

    So far, all I could manage to get is 1x^4(2)^0+4/1x^3(3). Is there an easier way to do this question without it becoming so messy/confusing?
    Define
    [nCr] = n!/[(n - r)!r!]

    Then
    (x + a)^n = [nC0]*x^n*a^0 + [nC1]*x^{n - 1}*a^1 + [nC2]*x^{n - 2}*a^2 + ... + [nC(n-1)]*x^1*a^{n - 1} + [nCn]*x^0*a^n

    In your case:
    (x + 2)^4 = [4C0]x^4 + [4C1](2)x^3 + [4C2](2)^2x^2 + [4C3](2)^3x + [4C4](2)^4

    Now
    [4C0] = 1
    [4C1] = 4
    [4C2] = 6
    [4C3] = 4
    [4C4] = 1

    So
    (x + 2)^4 = x^4 + 4*2x^3 + 6*4x^2 + 4*8x + 16

    = x^4 + 8x^3 + 24x^2 + 32x + 16

    -Dan
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jan 2007
    Posts
    26
    Quote Originally Posted by Logarithm Freak View Post
    Sn= (N/2)(A1 + An)
    Here's what I got:

    n = 48

    A1 = 4(1)-1
    A1 = 3

    A48 = 4(48)-1
    A48 = 191

    So,
    S = 48/2(3+191)
    S = 24(194)
    S = 4656

    Please let me know if I got this right.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,845
    Thanks
    320
    Awards
    1
    Quote Originally Posted by Mulya66 View Post
    A) "Find the sum of the arithmetic series"


    The more traditional way:

    Sum[4n - 1, 1, 48] = Sum[4n, 1, 48] - Sum[1, 1, 48]

    = 4*Sum[n, 1, 48] - Sum[1, 1, 48]

    Now,
    Sum[n, 1, N] = (N/2)(N + 1)
    Sum[1, 1, N] = N

    So
    Sum[4n - 1, 1, 48] = 4*Sum[n, 1, 48] - Sum[1, 1, 48]

    = 4*(48/2)(48 + 1) - 48

    = 4704 - 48

    = 4656
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Logarithm Freak View Post
    Sn= (N/2)(A1 + An)
    Don't just give an answer like this explain where it comes from, and for that
    matter what the variables/symbols represent.

    RonL
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: September 30th 2011, 01:40 PM
  2. Using Pascal's Triangle to Solve a sqrt Binomial
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: May 22nd 2010, 01:51 PM
  3. Pascal's triangle - Binomial coefficients
    Posted in the Algebra Forum
    Replies: 0
    Last Post: November 18th 2008, 02:47 AM
  4. Pascal's triangle
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: June 4th 2008, 09:37 PM
  5. Pascal's theorem and triangle
    Posted in the Statistics Forum
    Replies: 9
    Last Post: March 24th 2007, 10:17 PM

Search Tags


/mathhelpforum @mathhelpforum