Hi all

How I complete this question ..

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- Feb 14th 2010, 07:49 AMr-soyHow I complete this question ..
Hi all

How I complete this question .. - Feb 14th 2010, 07:56 AMe^(i*pi)
What are you trying to find? I can see no equals sign. Is it equal to 0?

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

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

$\displaystyle (2^k+1)+2^k(2^k+1) + 1 $

$\displaystyle 2^k + 1+ 2^{2k}+2^k+1$

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

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

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