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9 tháng 7 2023

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

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

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

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

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

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

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

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

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

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

9 tháng 7 2023

{3280}{6561}

27 tháng 8 2020

Bài làm:

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

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

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

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

=> \(A=\frac{3^8-1}{3^8.2}\)

27 tháng 8 2020

                          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 !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

30 tháng 3 2019

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

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

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

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

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

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

Vậy : ...

6 tháng 7 2015

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

21 tháng 7 2017

68890

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

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

Lấy 3A - A ta được :

\(2A=1-\dfrac{1}{6561}=\dfrac{6560}{6561}\Leftrightarrow A=\dfrac{6560}{6561}:2\)

\(\Leftrightarrow A=\dfrac{6560}{6561}.\dfrac{1}{2}=\dfrac{3280}{6561}\)

24 tháng 6 2023

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

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

\(2A=\dfrac{6560}{6561}\)

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

14 tháng 8 2016

1+3+9+27+....+2187+6561

14 tháng 8 2016

đặt A = 1+3+9+27+....+2187+6561 

=>A = 30 + 31 + 32 + 33 + .. . +37 + 38 

 3A = 31 + 3+ 3 + ... + 38 + 39

3A - A = (31 + 3+ 3 + ... + 38 + 39)-(30 + 31 + 32 + 33 + .. . +37 + 38 )

2A = 39 - 1 

A=\(\frac{3^9-1}{2}=\frac{19682}{2}=9841\)

14 tháng 8 2016

1+3+9+27+....+2187+6561

18 tháng 8 2018

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

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

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

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

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

18 tháng 8 2018

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= \(\frac{3280}{2187}\).