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Math Help - How I complete this question ..

  1. #1
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    How I complete this question ..

    Hi all

    How I complete this question ..
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  2. #2
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    Quote Originally Posted by r-soy View Post
    Hi all

    How I complete this question ..
    What are you trying to find? I can see no equals sign. Is it equal to 0?

    \frac{2^k-1}{2^k} + \frac{1}{2^k+1}


    1 - 2^{-k} + \frac{1}{2^k+1}

    (2^k+1)+2^k(2^k+1) + 1

    2^k + 1+ 2^{2k}+2^k+1

    2^{2k} + 2 \cdot 2^k + 2


    If this is equal to 0 then let u = 2^k so that the expression becomes u^2+2u+2. Use the quadratic formula to solve for u. Remember only values for which u > 0 are real solutions for k

    k = \log_2(u) = \frac{\ln(u)}{\ln(2)}
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