Cevap:
[tex]\frac{6}{2^{n}.2^{-2} } + \frac{10}{2^n.2^{1}} + \frac{12}{2^n.2^{2}} = 1\\\\\frac{6}{2^{n}.\frac{1}{4}} + \frac{10}{2^n}.\frac{1}{2} + \frac{12}{2^n.4} = 1\\\\\\\frac{6}{2^n}.4 + \frac{5}{2^n}+\frac{3}{2^n} = 1\\\\\\\frac{24}{2^{n}} + \frac{5}{2^{n}} + \frac{3}{2^n} = 1\\\\\\\frac{32}{2^{n}} = 1\\\\\\32 = 2^n\\\\2^5 = 2^n\\\\\ 5 = n[/tex]
