Results 1 to 4 of 4

Math Help - sigma notation

  1. #1
    Member
    Joined
    Sep 2005
    Posts
    84

    cubic sequence

    hello i have been asked to use sigma notation.

    i have a sequence here : 7,25,63,129,231,377,575.

    aparantly i can use a sigma notation for this.

    this is for a study which i am carrying out. above are sequences for total number of 3d squares for each sequences.

    i got a formula of 2d already which is 2n^2-2n+1 and i also got a formula for 3d using the formula an^3+bn^2+cn+d.

    but as a further part of the study i have to carry out sigma notation.

    i have attached the 2d and 3d pictures. for 2d you will reliase i have black squares please ignore that as that is a sepate part of the study. as the moment i am looking at the total number of squares.

    all help apreichated

    thank you.

    Attached Thumbnails Attached Thumbnails sigma notation-borders-proper.jpg   sigma notation-borders-2d.jpg  
    Last edited by rpatel; November 5th 2005 at 07:26 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Jun 2005
    Posts
    295
    Awards
    1
    I think this is what's going on here. You're looking at the number of cubes in a sequence of figures of which an example is shown. Let's look at the number of cubes in each layer: call that L(n). Each layer is composed of a number of rows each of which has an odd number of cubes in it. So L(n) = 1 + 3 + 5 + ... + (2n-1) + (2n+1) + (2n-1) + ... + 5 + 3 + 1 = sigma_{i=0}^{n} (2i+1) + sigma_{i=0}^{n-1} (2i+1). Now sigma_{i=0}^n i = n(n+1)/2 so sigma_{i=0}^n 2i+1 = n(n+1) + n+1 = (n+1)^2. So L(n) = (n+1)^2 + n^2 = 2n^2 + 2n + 1. The whole figure has F(n) = 1 + 5 + ... + L(n-1) + L(n) + L(n-1) + ... + 5 + 1 = sigma_{i=1}^n L(n) + sigma_{i=1}^{n-1} L(n) cubes in it. To evaluate this you need to know that sigma_{i=1}^n n^2 = n(n+1)(2n+1)/6.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2005
    Posts
    84
    how can i get the nth term of the sequence : 7,25,63,129,231,377,575

    i known the answer is 4n3+6n3+8n+3/3

    how can i get the answer exactly like this?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jun 2011
    Posts
    1

    Red face

    Quote Originally Posted by rpatel View Post
    how can i get the nth term of the sequence : 7,25,63,129,231,377,575

    i known the answer is 4n3+6n3+8n+3/3

    how can i get the answer exactly like this?
    f(n)=an^3+bn^2+cn+d
    You need to find the differences until you get the common difference.

    1 7, 25, 63, 129, 231, 377, 575
    6 18 38 66 102 146 198 1d
    12 20 28 36 44 52 2d
    8 8 8 8 8 3d
    NB the underlined numbers are for reversing.
    therefore to find a: 2a=3diff=8
    therefore a = 4
    To find d you need to find f(0) by revesing. therefore d= 1
    to find b and c you need two equations, therefore the easiest is f(1) and f(2) by solving these equations silmultaneously you will find the desired values i.e b=-6 and c=8.
    Last edited by mr fantastic; June 2nd 2011 at 07:17 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. sigma notation
    Posted in the Algebra Forum
    Replies: 3
    Last Post: June 24th 2010, 08:35 AM
  2. Sigma Notation Help
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: February 26th 2010, 11:15 PM
  3. Sigma notation help
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 29th 2009, 06:07 PM
  4. sigma notation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 14th 2009, 03:17 PM
  5. Sigma notation
    Posted in the Algebra Forum
    Replies: 5
    Last Post: January 24th 2008, 02:30 AM

Search Tags


/mathhelpforum @mathhelpforum