C =           \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\) + \(\dfrac{1}{128}\)

  2\(\times\)C =   1 +  \(\dfrac{1}{2}\)  + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\) 

2 \(\times\) C - C =   1 -  \(\dfrac{1}{128}\)

       C       = \(\dfrac{127}{128}\)

Mk có cách giải khác nè

1/4+1/8+1/16+1/32+1/64+1/128

= 1/2-1/4+1/4-1/8+1/8-1/16+1/16-1/32+1/32-1/64+1/64-1/128

= 1/2-1/128

= 63/128

63/128

=\(\frac{32}{128}\)+\(\frac{16}{128}\)+\(\frac{8}{128}\)+\(\frac{4}{128}\)+\(\frac{1}{128}\)=\(\frac{32+16+8+4+1}{128}\)=\(\frac{61}{128}\)

a: \(A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^7\)

=>\(2\cdot A=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^6\)

=>\(2A-A=1-\left(\dfrac{1}{2}\right)^7=1-\dfrac{1}{128}=\dfrac{127}{128}\)

=>\(A=\dfrac{127}{128}\)

b: \(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{10\cdot11}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{10}-\dfrac{1}{11}\)

\(=1-\dfrac{1}{11}=\dfrac{10}{11}\)

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)

\(=1-\frac{1}{2}+...+\frac{1}{128}=1-\frac{1}{128}=\frac{127}{128}\)

b: A=1/3+1/9+...+1/3^10

=>3A=1+1/3+...+1/3^9

=>A*2=1-1/3^10=(3^10-1)/3^10

=>A=(3^10-1)/(2*3^10)

c: C=3/2+3/8+3/32+3/128+3/512

=>4C=6+3/2+...+3/128

=>3C=6-3/512

=>C=1023/512

d: A=1/2+...+1/256

=>2A=1+1/2+...+1/128

=>A=1-1/256=255/256

*Tham khảo 

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)

đặt A = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\) 

\(A\cdot2=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)

\(A\cdot2-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\) \(-\) \(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\right)\)

\(A=\) \(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\frac{1}{32}-\frac{1}{64}-\frac{1}{128}\)

\(A=1-\frac{1}{128}\)

\(A=\frac{127}{128}\)

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)

Ta lấy\(\frac{1}{128}\)là MSC. Ta tính được \(\frac{64}{128}+\frac{32}{128}+\frac{16}{128}+\frac{8}{128}+\frac{4}{128}+\frac{2}{128}+\frac{1}{128}\)

Kết quả bằng \(\frac{127}{128}\)

127/128

1/2 + 1/4 + 1/8 + 1/16 +1/32 + 1/64 + 1/128

=1-1/2+1/2-1/4+1/4-1/8+...+1/64+1/128

=1-1/128

=127/128

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{64}-\frac{1}{128}\)

\(=1-\frac{1}{128}\)

\(=\frac{127}{128}\)

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{64}-\frac{1}{128}\)

\(=1-\frac{1}{128}\)

\(=\frac{127}{128}\)

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=\(\frac{63}{128}\)

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)\(+\frac{1}{64}+\frac{1}{128}\)

\(=\frac{1}{1.2}+\frac{1}{2.2}+\frac{1}{2.3}\)\(+\frac{1}{2.4}+\frac{1}{2.8}+\frac{1}{4.8}+\frac{1}{8.8}+\frac{1}{8.16}\)

\(=\frac{1}{1}-\frac{1}{16}=\frac{16}{16}-\frac{1}{16}=\frac{15}{16}\)