1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook