A=1+1/2+1/3+1/6+(1/12+1/15+1/20+1/30)+1/35

=71/35+7/30=95/42

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

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

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

\(B=\frac{6}{1\cdot3}+\frac{6}{3\cdot5}+\cdot\cdot\cdot+\frac{6}{97\cdot99}\)

\(\Rightarrow B=3\cdot\left(\frac{2}{1\cdot3}+\cdot\cdot\cdot+\frac{2}{97\cdot99}\right)\)

\(\Rightarrow B=3\cdot\left(1-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{97}-\frac{1}{99}\right)\)

\(\Rightarrow B=3\cdot\left(1-\frac{1}{99}\right)\)

\(\Rightarrow B=3\cdot\frac{98}{99}\)

\(\Rightarrow B=\frac{98}{33}\)

\(A=\frac{1}{2}+\frac{1}{6}+\cdot\cdot\cdot+\frac{1}{42}\)

\(\Rightarrow A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdot\cdot\cdot+\frac{1}{6\cdot7}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{6}-\frac{1}{7}\)

\(\Rightarrow A=1-\frac{1}{7}\)

\(\Rightarrow A=\frac{6}{7}\)

a)\(A=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)

\(\frac{1}{2}xA=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\)

\(\frac{1}{4}xA=\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}+\frac{1}{384}\)

\(\frac{1}{4}xA-\frac{1}{2}xA=\frac{1}{3}-\frac{1}{384}\)

\(\frac{1}{4}xA=\frac{127}{384}\)

\(A=\frac{127}{384}:\frac{1}{4}\)

\(A=\frac{127}{96}\)

\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)

\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)

\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)

\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)

\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)

\(=9-\left(1-\frac{1}{10}\right)\)

\(=9-\frac{9}{10}=\frac{81}{10}\)

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

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

C=1-1/128

C=127/128

D=1/2+1/6+1/12+1/20+1/30+1/42

D=1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7

D=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7

D=1/1-1/7

D=6/7

\(a,\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)

\(=1-\frac{1}{7}\)

\(=\frac{6}{7}\)

\(b,\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)

Ta có :

\(\frac{1}{2}=1-\frac{1}{2}\)

\(\frac{1}{4}=\frac{1}{2}-\frac{1}{4}\)

\(\frac{1}{8}=\frac{1}{4}-\frac{1}{8}\)

\(\frac{1}{16}=\frac{1}{8}-\frac{1}{16}\)

\(\frac{1}{32}=\frac{1}{16}-\frac{1}{32}\)

Thay vào ta có :

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

\(=1-\frac{1}{32}\)

\(=\frac{31}{32}\)

\(c,\)\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\)

Ta có :

\(\frac{1}{2}=1-\frac{1}{2}\)

\(\frac{1}{4}=\frac{1}{2}-\frac{1}{4}\)

...................

\(\frac{1}{256}=\frac{1}{128}-\frac{1}{256}\)

Thay vào ta có :

\(=\)\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{128}-\frac{1}{256}\)

\(=1-\frac{1}{256}\)

\(=\frac{255}{256}\)