Posers and Puzzles

Posers and Puzzles

  1. Joined
    21 Apr '05
    Moves
    54
    22 May '05 06:42
    Sum the series:

    1*2*3-2^3+3*4*5-4^3+...+9999*10000*10001-10000^3.
  2. Joined
    29 Feb '04
    Moves
    22
    22 May '05 10:19
    -250500
  3. Joined
    12 Mar '03
    Moves
    37890
    22 May '05 10:451 edit
    - 25 005 000

    edit: but of course, it is not real 😀
  4. Joined
    29 Feb '04
    Moves
    22
    22 May '05 11:071 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. Joined
    21 Apr '05
    Moves
    54
    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. Joined
    29 Feb '04
    Moves
    22
    17 Jun '05 12:101 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


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