Question

2^2008 - 2^2007 - 2^2006 + 2^2005 = 9^k . 2^2005 por favooor....

Answer 1

Olá Lucas!

$\\ 2^{2008} - 2^{2007} - 2^{2006} + 2^{2005} = 9^k \cdot 2^{2005} \\\\ 2^{2005}(2^3 - 2^2 - 2^1 + 2^0) = (3^2)^k \cdot 2^{2005} \\\\ (8 - 4 - 2 + 1) = 3^{2k} \\\\ 3 = 3^{2k} \\\\ 2k = 1 \\\\ \boxed{k = \frac{1}{2}}$

 Espero ter ajudado!