\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}+\frac{1}{6561}\)

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

\(3A-A=\left[1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\right]-\left[\frac{1}{3}+\frac{1}{9}+...+\frac{1}{6561}\right]\)

\(2A=1-\frac{1}{6561}=\frac{6560}{6561}\)

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

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

Vậy : ...

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

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

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

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

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

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

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

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

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

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

{3280}{6561}

\(\)

S x 3 = 3 + 1 + 1/3 + 1/9 + 1/27 + ..................... + 1/729

S x 3 – S = 3 – 1/2187 = 6560/2187

Vậy S =  6560/2187 : 2 = 6560/4374

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

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

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

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

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

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

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

=>3S= 3+ 1+ \(\frac{1}{3}\)+...+ \(\frac{1}{3^5}\)+ \(\frac{1}{3^6}\).

=> 3S- S=( 3+ 1+ \(\frac{1}{3}\)+...+ \(\frac{1}{3^5}\)+ \(\frac{1}{3^6}\))-( 1+ \(\frac{1}{3}\)+ \(\frac{1}{3^2}\)+...+ \(\frac{1}{3^6}\)+ \(\frac{1}{3^7}\)).

=> 2S= 3- \(\frac{1}{3^7}\).

=> 2S= 3- \(\frac{1}{2187}\).

=> 2S= \(\frac{6560}{2187}\).

=> S= \(\frac{6560}{2187}\): 2.

=> S= \(\frac{3280}{2187}\).

Vậy S= \(\frac{3280}{2187}\).

C = 1 + 2 + 4 + 8 + ... + 1024

2 x C = 2 + 4 + 8 + ... + 1024 + 2048

 2 x C - C = C = (2 + 4 + 8 + ... + 1024 + 2048) - (1 + 2 + 4 + 8 + ... + 1024) = 2048 - 1 = 2047

D = 1 + 3 + 9 + 27 + ... + 2187

3 x D = 3 + 9 + 27 + ... + 2187 + 6561

3 x D - D = 2 x D = (3 + 9 + 27 + ... + 2187 + 6561) - (1 + 3 + 9 + 27 + ... + 2187) = 6561 - 1 = 6560

D = 6560 : 2 = 3280

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

Nhớ trả lời cho mình

Mình giúp bạn nè  

Ta có:

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

\(\Rightarrow3A=3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)

\(\Rightarrow3A-A=\left(3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\right)-\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}\right)\)

\(\Rightarrow2A=3-\frac{1}{2187}=\frac{6561}{2187}-\frac{1}{2187}=\frac{6560}{2187}\)

\(\Rightarrow A=\frac{6560}{2187}:2=\frac{3280}{2187}\)

cảm ơn bạn nhiều lắm

ko có j