\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{64}-\frac{1}{128}\)
\(=1-\frac{1}{128}\)
\(=\frac{127}{128}\)
A=\(\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)=\frac{1}{2}\left(1+A-\frac{1}{2}-\frac{1}{128}\right)\)
2A=\(A+\frac{1}{2}-\frac{1}{128}=A+\frac{63}{128}\)
=> A=\(\frac{63}{128}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)\(+\frac{1}{64}+\frac{1}{128}\)
\(=\frac{1}{1.2}+\frac{1}{2.2}+\frac{1}{2.3}\)\(+\frac{1}{2.4}+\frac{1}{2.8}+\frac{1}{4.8}+\frac{1}{8.8}+\frac{1}{8.16}\)
\(=\frac{1}{1}-\frac{1}{16}=\frac{16}{16}-\frac{1}{16}=\frac{15}{16}\)