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Ap/GP Question

  1. Apr 8, 2005 #1
    Hi everyone,

    This is my first post here :smile: Anyway I have problems solving this question wonder anyone could help give me some clues as to how to go about it. Here goes:

    The positive integers are bracketed as follows,

    (1), (2,3), (4,5,6,7), (8,9,10,11,12,13,14,15), ...

    No. of integers in the rth bracket is 2^r-1. State stae first term and last term in the nth bracket. Hence, show that the sum of the terms in the first n brackets is 2^n-1[(2^n)]-1].

    Thanks very much.. I have absolutely no clue..

  2. jcsd
  3. Apr 8, 2005 #2


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    Homework Helper

    Start by finding how many numbers are in the first n brackets, which is just the sum of the first n powers of 2 (use the geometric sum formula). This will let you know the first and last number in each bracket. Then, to find the sum, use the sum of the first n positive integers, and plug in what you just found for n. If you don't know the sum of the first n integers, the easiest way to find it is to use a telescoping sum:

    [tex] \sum_{k=1}^{n} (k+1)^2 - k^2 [/tex]

    If you were to write this out term by term, you'd see each term cancels except the first and last. Now that you have this sum, rewrite it as:

    [tex] \sum_{k=1}^{n} k^2+ 2k + 1 - k^2 = \sum_{k=1}^{n} 2k + 1[/tex]

    from which the result should follow pretty easily.
    Last edited: Apr 8, 2005
  4. Apr 9, 2005 #3
    I got it.. thanks status.
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