giải giúp mình bài toan: A=1+1/3+1/9+1/27+...+1/729
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a x 3 = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243
a x 3 - a = (1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243) - (1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729)
a x 2 = 1 - 1/729
a x 2 = 728/729
a = 728/729 : 2 = 364/729

Mình nghĩ là đề sai. Các phân số đều có mẫu là lũy thừa của 3 vì thế 1/143 phải là 1/243 chứ
= ( 1 +729 + 1/143 ) + ( 1/3 + 1/9 + 1/27 + 1/81 )
= ( 730 + 1/143 ) + ( 27/81 + 9/81 + 3/81 + 1/81 )
= ( 730 + 1/143 ) + 40/81
= 104391/143 + 40/81
= 730, 5008202

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
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)
B x 3 = 1 + 1/3 + 1/9 +.........+1/243 +1/729
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

\(S=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(3\times S=3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(3\times S-S=\left(3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)-\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\right)\)
\(2\times S=3-\frac{1}{729}\)
\(S=\frac{1093}{729}\)

`Answer:`
\(\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{9}\right)+\left(x+\frac{1}{27}\right)+...+\left(x+\frac{1}{729}\right)=\frac{4209}{729}\)
\(\Leftrightarrow\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{3^2}\right)+\left(x+\frac{1}{3^3}\right)+...+\left(x+\frac{1}{3^6}\right)=\frac{4209}{729}\)
\(\Leftrightarrow6x+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^6}\right)=\frac{4209}{729}\text{(*)}\)
Đặt \(N=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^6}\)
\(\Leftrightarrow3N=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\)
\(\Leftrightarrow3N-N=\left(1+\frac{1}{3}+\frac{1}{3^2}+..+\frac{1}{3^5}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^6}\right)\)
\(\Leftrightarrow2N=1-\frac{1}{3^6}\)
\(\Leftrightarrow2N=\frac{728}{729}\)
\(\Leftrightarrow N=\frac{364}{729}\)
\(\text{(*)}\Leftrightarrow6x+\frac{364}{729}=\frac{4209}{729}\)
\(\Leftrightarrow6x=\frac{3845}{729}\)
\(\Leftrightarrow x=\frac{3845}{4374}\)

1 + 1/3 + 1/9+1/27+1/81+1/243+1/729
=1+1-1/3+1/3-1/9+1/9-1/27-1/27-1/81+1/81-1/243
= 2 - 1/243
=485/243