A=1/3+1/9+1/27+...+1/2187
   =1/3+1/3^2+1/3^3+...+1/3^7
-->3A=1+1/3+1/3^2+...+1/3^6
-->3A-A=(1+1/3+1/3^2+...+1/3^6) - (1/3+1/3^2+1/3^3+...+1/3^7)
-->2A=1- 1/3^7
-->A=1093/2187

ko hiểu có thể giải lại dc ko bạn

pn ơi cho mk hỏi các pn học mũ (x²) chua

\(S=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\)

\(=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)

\(\Rightarrow\)\(3S=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\)

\(\Rightarrow\)\(3S-S=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)\)

\(\Rightarrow\)\(2S=1-\frac{1}{3^7}\)

\(\Rightarrow\)\(S=\frac{1-\frac{1}{3^7}}{2}\)

\(S=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)

\(3S=1+\frac{1}{3}+...+\frac{1}{3^6}\)

\(3S-S=\left(1+\frac{1}{3}+...+\frac{1}{3^6}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)\)

\(2S=1-\frac{1}{3^7}\)

\(S=\frac{1-\frac{1}{3^7}}{2}\)

\(S=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\)

\(S=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)

\(\Rightarrow3S=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\)

\(\Rightarrow3S-S=1-\frac{1}{3^7}\)

\(2S=1-\frac{1}{3^7}\)

\(\Rightarrow S=\frac{1-\frac{1}{3^7}}{2}\)

Gọi tong trên là A

\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{81}+\frac{1}{243}+\frac{1}{7129}+\frac{1}{2187}\)

\(3A=\frac{1}{3}+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{729}\)

\(3A-A=\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\right)-\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}\right)\)

\(2A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}-\frac{1}{3}-\frac{1}{9}-\frac{1}{27}-\frac{1}{81}-\frac{1}{243}-\frac{1}{729}-\frac{1}{2187}\)

\(2A=1-\frac{1}{2187}\)

\(2A=\frac{2186}{2187}\)

\(A=\frac{2186}{2187}:2\)

\(A=\frac{1093}{2187}\)

Vậy tổng A = \(\frac{1093}{2187}\)

\(3y=3\cdot\frac{1}{1}+3\cdot\frac{1}{3}+3\cdot\frac{1}{9}+...+3\cdot\frac{1}{729}+3\cdot\frac{1}{2187}\)

     \(=3+\frac{1}{1}+\frac{1}{3}...+\frac{1}{729}\)

=> \(3y-y=3+\frac{1}{1}+\frac{1}{3}+..+\frac{1}{729}-\frac{1}{1}-\frac{1}{3}-...-\frac{1}{2187}\)

<=> 2y = 3- 1/2187

=> y = \(\frac{3-\frac{1}{2187}}{2}\)

\(\text{Đ}\text{ặt} A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}\)

\(\Rightarrow2187A=2187+729+243+81+27+9+3+1\)

\(\Leftrightarrow2187A=3280\)

\(\Leftrightarrow A=\frac{3280}{2187}\)

Chắc chắn 100% luôn

\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)

\(\Rightarrow\frac{1}{3}B=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^8}\)

\(\Rightarrow B-\frac{1}{3}B=\frac{1}{3}-\frac{1}{3^8}\Rightarrow\frac{2}{3}B=\frac{3^7-1}{3^8}\Rightarrow B=\frac{3\left(3^7-1\right)}{2.3^8}\)

Ta có : 

\(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\)

\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)

\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\)

\(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\right)\)

\(2B=1-\frac{1}{3^7}\)

\(2B=\frac{3^7-1}{3^7}\)

\(B=\frac{3^7-1}{3^7}:2\)

\(B=\frac{3^7-1}{2.3^7}\)

Vậy \(B=\frac{3^7-1}{2.3^7}\)

Chúc bạn học tốt ~

\(=[3\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{81}+...+\frac{1}{2187}\right)-\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{81}+...+\frac{1}{2187}\right)]:2\)

\(=\left(1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{729}-\frac{1}{3}-\frac{1}{9}-\frac{1}{81}-...-\frac{1}{2187}\right):2\)

\(=\left(1-\frac{1}{2187}\right):2=\frac{2186}{2187}.\frac{1}{2}=\frac{1093}{2187}\)

Đặt B = 1/3 + 1/9 + 1/27 + 1/81 +1/243 + 1/729 + 1/2187

B x 3 = 3 x ( 1/3 + 1/9 +.......+ 1/729 + 1/2187)

        =  1 + 1/3 + 1/9 +.........+1/243 +1/729 

Lấy B x 3 - B ta có :

B x 3 - B = 1 + 1/3 +1/9+ .........+1/243 + 1/729 - 1/3 + 1/9 +.........+1/729 +1/2187

B x (3 - 1)= 1 - 1/2187

B x  2     =   2186/2187

B       = 2186/2187 : 2 = 1093/2187