Please turn on javascript in your browser to play chess.
Posers and Puzzles

Posers and Puzzles

  1. 22 May '05 06:42
    Sum the series:

    1*2*3-2^3+3*4*5-4^3+...+9999*10000*10001-10000^3.
  2. 22 May '05 10:19
    -250500
  3. 22 May '05 10:45 / 1 edit
    - 25 005 000

    edit: but of course, it is not real
  4. 22 May '05 11:07 / 1 edit
    Originally posted by Mephisto2
    - 25 005 000

    edit: but of course, it is not real
    Yes, I was going to add '...provided negative numbers exist'.
  5. 17 Jun '05 11:53
    Originally posted by Mephisto2
    - 25 005 000

    edit: but of course, it is not real
    I was wondering if someone smarter than me can explain how they got an answer.

    All i have is;

    If I split the entire series into two parts

    (1*2*3+3*4*5+.........)-(2^3+4^3+.......)

    The first part can be written as Summation[n(n+1)(n+2)]

    Summation (n^3 + 3(n^2) + 2n)

    For the second part take 2^3 common
    2^3(1+2^3 + 3^3......)
    2^3 Summation(n^3)
  6. 17 Jun '05 12:10 / 1 edit
    Oops, looks like I summed up to 500 rather than 5000.

    The general term is
    (2n-1)(2n)(2n+1) - (2n)^3
    = -2n

    Summing between 1 and 5000 gives -25005000