Results 1 to 2 of 2

Math Help - Factoring

  1. #1
    Member
    Joined
    Sep 2007
    Posts
    222

    Question Factoring

    Bit confused on how this:
    T6+2T6+3T6+...+(n-1)T6+nT6
    factored gives you:
    T6[1/2(n^2+n)]

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Mar 2010
    Posts
    993
    Thanks
    244
    I'm not sure what the notation T6 means, but it looks like you can divide it out of the equation, giving:
    1+2+3+...+(n-1)+n = 1/2(n^2+n) (which is equal to n(n+1)/2)

    If n is even, you can group the left hand side like this:
    (1+n) + (2+(n-1)) + (3+(n-2)) + ... + ((n/2)+((n/2)+1)
    and there are n/2 terms which are all equal to n+1, so the sum is n(n+1)/2.

    If n is odd, you can do the same thing without the last term:
    1+2+3+...+(n-1)+n = (1+2+3+...+(n-1)) + n = (n-1)(n)/2 + n
    which simplifies to (n^2-n)/2 + 2n/2 = (n^2-n+2n)/2 = (n^2+n)/2 = n(n+1)/2
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. need help with factoring
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 7th 2010, 10:47 AM
  2. factoring help
    Posted in the Algebra Forum
    Replies: 2
    Last Post: February 4th 2010, 07:44 PM
  3. Is this factoring or something?
    Posted in the Algebra Forum
    Replies: 4
    Last Post: February 1st 2010, 06:54 PM
  4. Replies: 2
    Last Post: August 22nd 2009, 10:57 AM
  5. Replies: 3
    Last Post: November 5th 2006, 11:02 PM

Search Tags


/mathhelpforum @mathhelpforum