G=[tex]\frac{1}{3} + \frac{1}{6} + \frac{1}{12} + ... +\frac{1}{1536}[/tex]
[tex]\dpi{100} G=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{1536}\\=\frac{1}{3}+\frac{1}{2}.\frac{1}{3}+\frac{1}{2^2}.\frac{1}{3}+...+\frac{1}{2^9}.\frac{1}{3}\\=\frac{1}{3}.(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9})[/tex]
Đặt
[tex]\dpi{100} D=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\\2D=2+1+\frac{1}{2}+...+\frac{1}{2^8}\\2D-D=(2+1+\frac{1}{2}+...+\frac{1}{2^8})-(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9})\\D=2-\frac{1}{2^9}\\D=\frac{2^{10}}{2^{9}}-\frac{1}{2^9}\\=\frac{2^{10}-1}{2^9}\\=\frac{1023}{512}[/tex]
[tex]\dpi{100} G=\frac{1}{3}.D=\frac{1}{3}.\frac{1023}{512}=\frac{341}{512}[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
