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 \(V=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+...+\dfrac{1}{729}+\dfrac{1}{2187}\)

\(\Rightarrow3V=3.\left(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+...+\dfrac{1}{729}+\dfrac{1}{2187}\right)\)

\(\Rightarrow3V=1+\left(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+...+\dfrac{1}{729}\right)\)

\(\Rightarrow3V=1+V-\dfrac{1}{2187}\)

\(\Rightarrow2V=1-\dfrac{1}{2187}\)

\(\Rightarrow V=\dfrac{1093}{2187}\).

1093/2187 nha bạn

A= 1/3 + 1/9 + 1/27 + 1/81 +...+1/729 + 1/2187

3A = 1 + 1/3 + 1/9 + 1/27 + 1/81 +...+1/729 

=> 3A - A = 1 - 1/2187

lấy MS chung là 2187, ta có:

        729 + 243 + 81 + 9 + 3 + 1                         

       ________________________  =      1066/2187

                      2187

1066/2187.

Bài làm:

Ta có: \(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}+\frac{1}{6561}=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}+\frac{1}{3^8}\)

=> \(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}+\frac{1}{3^7}\) 

=> \(3A-A=\left(1+\frac{1}{3}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)\)

<=> \(2A=1-\frac{1}{3^8}=\frac{3^8-1}{3^8}\)

=> \(A=\frac{3^8-1}{3^8.2}\)

                          Bài làm :

Ta có :

\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{6561}\)

\(\Rightarrow3\times A=\frac{1\times3}{3}+\frac{1\times3}{9}+\frac{1\times3}{27}+...+\frac{1\times3}{6561}\)

\(3\times A=1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{729}+\frac{1}{2187}\)

\(3\times A=1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{729}+\frac{1}{2187}+\left(\frac{1}{6561}-\frac{1}{6561}\right)\)

\(3\times A=1+\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{729}+\frac{1}{2187}+\frac{1}{6561}\right)-\frac{1}{6561}\)

\(3\times A=1+A-\frac{1}{6561}\)

\(\Rightarrow2\times A=1-\frac{1}{6561}\)( Trừ bỏ A ở cả 2 vế )

\(2\times A=\frac{6560}{6561}\)

\(A=\frac{6560}{6561}\div2=\frac{3280}{6561}\)

Vậy A=3280/6561

Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

\(B=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+...+\dfrac{1}{2187}+\dfrac{1}{6561}\)

\(3B=3\cdot\left(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+...+\dfrac{1}{6561}\right)\)

\(3B=1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{729}+\dfrac{1}{2187}\)

\(3B-B=\left(1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{2187}\right)-\left(\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{6561}\right)\)

\(2B=\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{9}-\dfrac{1}{9}\right)+...+\left(1-\dfrac{1}{6561}\right)\)

\(2B=0+0+...+1-\dfrac{1}{6561}\)

\(2B=1-\dfrac{1}{6561}\)

\(B=\left(1-\dfrac{1}{6561}\right):2\)

\(B=\dfrac{6560}{6561}:2\)

\(B=\dfrac{3280}{6561}\)

{3280}{6561}

\(\)

ta có :

= ( 1 + 59049 ) + ( 3 + 2187 ) + ( 9 + 6561 ) + ( 27 + 243 ) + ( 81 + 729 )

= 59050 + 2190 + 6570 + 270 + 810

= 59050 + ( 2190 + 810 ) + 6570 + 270

= 59050 + 3000 + 6570 + 270

= 59050 + ( 3000 + 6570 ) + 270

= 59050 + 9570 + 270

= 68620 + 270

= 68890

68890

Kết quả là 68890

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