\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)

=>\(2A=1+\frac{1}{2}+...+\frac{1}{2^8}\)

=>\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)

                       \(=1-\frac{1}{2^9}=\frac{511}{512}\)

\(B=\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}+\frac{1}{972}\)

=>\(3B=\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}\)

=>\(3B-B=\left(\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+...+\frac{1}{324}\right)-\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+...+\frac{1}{972}\right)\)

=>\(2B=\frac{3}{4}-\frac{1}{972}=\frac{182}{243}\)

=>\(B=\frac{182}{243}:2=\frac{91}{243}\)

\(A=\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}\)

\(3A=3\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}+\frac{1}{972}\right)\)

\(3A=\frac{3}{4}+\frac{3}{12}+\frac{3}{36}+\frac{3}{108}+\frac{3}{324}+\frac{3}{927}\)

\(3A=\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}\)

\(2A=3A-A\)

\(2A=\left(\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}\right)-\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}+\frac{1}{972}\right)\)

\(2A=\frac{3}{4}-\frac{1}{927}\)

\(2A=\frac{729-1}{972}=\frac{728}{972}=\frac{182}{243}\)

\(A=\frac{182}{243}:\frac{1}{2}\)

\(A=\frac{364}{243}\)

A=1/4+1/12+1/36+1/108+1/324+1/972

=243/972+81/972+27/972+9/972+3/972+1/972

=364/972

=91/243

\(B=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}\)

\(B=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{20}\right)+\left(1-\frac{1}{30}\right)+\left(1-\frac{1}{42}\right)+\left(1-\frac{1}{56}\right)\)

\(B=1.9-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{56}\right)\)

\(B=9-\left[\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{56}\right)\right]\)

\(B=9-\left(1-\frac{1}{56}\right)\)

\(B=9-\frac{55}{56}\)

\(B=\frac{449}{56}\)

Đặt A= 1/4+1/12+1/36+1/108+1/324 +1/972

            =1/4.(1+1/3+1/9+1/27+1/81+1/243)

            =1/4(1+1/3 +1/3^2 +1/3^3 +1/3^4 +1/3^5)

             gọi M=1+1/3+1/3^2+1/3^3+1/3^4+1/3^5

                     3M=3+1+1/3+1/3^2+1/3^3+1/3^4

                     3M-M=(3+1+1/3+1/3^2+1/3^3+1/3^4)-(1+1/3+1/3^2+1/3^3+1/3^4+1/3^5)

                         2M=3-1/3^5

                         M=3/2-1/3^5.2  

                A=1/4.(3/2-1/3^5.2)

                A=3/8-1/1944=91/243

                  Vậy .......

                Chúc bạn học tốt .....^-^      

\(3C=1+\frac{1}{3}+.....+\frac{1}{3^{2015}}\)

\(\Rightarrow3C-C=2C=\left(1+\frac{1}{3}+.....+\frac{1}{3^{2014}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}.....+\frac{1}{3^{2015}}\right)=1-\frac{1}{3^{2015}}\)

\(\Rightarrow C=\frac{3^{2015}-1}{3^{2015}.2}\)

\(D=4\left(1+\frac{1}{3}+....+\frac{1}{3^5}\right)\)

\(\Rightarrow3D=4\left(3+1+....+\frac{1}{3^4}\right)\)

\(\Rightarrow3D-D=2D=4\left(3+1+....+\frac{1}{3^4}\right)-4\left(1+\frac{1}{3}+....+\frac{1}{3^5}\right)\)

\(\Rightarrow2D=4\left(3-\frac{1}{3^5}\right)\Rightarrow D=2\left(3-\frac{1}{3^5}\right)\)

A=511/512

B=91/243

A=511/512

B=91/243

A=511/512

B=91/243

 Đặt tổng trên là A ta có 
A/3=1/12+1/36+1/108+ +1/2916+1/8748 
A-A/3=2A/3= 
=1/4-1/8748=2186/8748 
=>A=1093/1916

k mk nhá

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91/243

91/243 đó em

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