# How I complete this question ..

• February 14th 2010, 07:49 AM
r-soy
How I complete this question ..
Hi all

How I complete this question ..
• February 14th 2010, 07:56 AM
e^(i*pi)
Quote:

Originally Posted by r-soy
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

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