1/128+2/128+4/128+8/128+16/128+32/128+64/128=127/128
k di
\(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)đề là như vậy mới đúng
=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}+\frac{1}{64}+\frac{1}{128}\)
=\(1-\frac{1}{128}=\frac{127}{128}\)
= \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+....+\frac{1}{64}-\frac{1}{128}\)
=\(\frac{1}{2}-\frac{1}{128}\)
=\(\frac{63}{128}\)