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Math Help - help with math question

  1. #1
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    help with math question

    help me with this please:
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  2. #2
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    Quote Originally Posted by tukilala View Post
    help me with this please:
    That is: prove that \sum_{i= 0}^{log_2(n-1)} 2^i= n-1

    Did your try this with n= 1, 2, 3, etc. first? If n= 1, n-1= 0 and log_2(0) is not defined. if n= 2, log_2(1)= 0 so that sum is 2^0= 1= 2-1. If n= 3, log_2(2)= 1 so the sum is 2^0+ 2^1= 1+ 2= 3\ne 3-1.

    In order for that sum to make sense, n-1 must be a power of 2 so n must be one more than a power of 2. For example, if n= 5, n-1= 4 and log_2(n-1)= log_2(4)= 2 so the sum is 2^0+ 2^1+ 2^2= 1+ 2+ 4= 7\ne 5- 1.

    You can't prove this- it is not true. What can be proven is that if n is a power of 2 (1, 2, 4, 8, etc.) then \sum_{i=0}^{log_2(n)} 2^i= 2n-1 or, more simply, \sum_{i=0}^n 2^i= 2^{n+1}-1.
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  3. #3
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    Quote Originally Posted by tukilala View Post
    help
    If, for example, n= 5, then ln_2(n-1)= ln_2(4)= 2 so the sum is from 0 to 2: 1+ 2+ 4= 7 which is NOT equal no to n-1= 4. You can't prove it, it's not true.
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