TM
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
KL
1
LN
17 tháng 3 2016
A=$\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+......+\frac{1}{59049}$
3A=$\frac{1}+frac{1}{3}+\frac{1}{9}+\frac{1}{27}+......+\frac{1}{19683}$
3A-A=2A=1-1/59049=59048/59049
A=59048/118098
15 tháng 8 2015
I = 1 + 3^1 + 3^2 + ...+3^10
3I = 3^1 + 3^2 + ... + 3^11
3I - I = 3^1 + 3^2 + ... + 3^11 - 1 - 3^1 - 3^2 - ... - 3^10
2I = 3^11 - 1
I = \(\frac{3^{11}-1}{2}=\frac{177147-1}{2}=\frac{177146}{2}=88573\)
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= \(\frac{3280}{2187}\).
MT
0
NH
0
5 tháng 10 2016
Đặt \(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\)
\(3A=3\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\right)\)
\(3A=3+1+...+\frac{1}{3^4}\)
\(3A-A=\left(3+1+...+\frac{1}{3^4}\right)-\left(1+\frac{1}{3}+...+\frac{1}{3^5}\right)\)
\(2A=3-\frac{1}{3^5}\)
\(A=\frac{3-\frac{1}{3^5}}{2}\)
5 tháng 10 2016
Đặt \(S=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(S=1+\frac{1}{1\times3}+\frac{1}{3\times3}+\frac{1}{9\times3}+\frac{1}{27\times3}+\frac{1}{81\times3}\)
\(S\times3=\left(1+\frac{1}{1\times3}+\frac{1}{3\times3}+\frac{1}{9\times3}+\frac{1}{27\times3}+\frac{1}{81\times3}\right)\times3\)
\(S\times3=3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)
Xét: \(S\times3-S=\left(3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\right)-\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(S\times2=3-\frac{1}{243}\)
\(S\times2=\frac{728}{243}\)
\(S=\frac{728}{243}\div2\)
\(S=\frac{364}{243}\)
Vậy \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}=\frac{364}{243}\)
MG
11 tháng 4 2018
=182.\(\orbr{\begin{cases}1.\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)\\2.\left(\frac{1}{2}+\frac{1}{9}+\frac{1}{27}\right)\end{cases}}:\frac{4.\left(\frac{1}{7}+\frac{1}{9}-\frac{1}{343}\right)}{1.\left(\frac{1}{3}+\frac{1}{49}-\frac{1}{343}\right)}:\frac{91}{80} \)
=.\(182.\left(\frac{1}{2}:\frac{4}{1}\right).\frac{91}{80}\)
=\(182.\frac{1}{8}.\frac{91}{80}\)
=.\(182.\frac{91}{640}\)
=\(\frac{8281}{320}\)
B
11 tháng 4 2018
\(=182.\left[\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2.\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}:\frac{4.\left(1-\frac{1}{7}+\frac{1}{9}-\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{9}-\frac{1}{343}}\right]:\frac{919191}{808080}\)
\(=182.\frac{1}{8}.\frac{808080}{919191}=\frac{182}{8}.\frac{80}{91}=20\)
TH
3
16 tháng 3 2016
Đặt A= 1/3+1/9+1/27+1/81+1/243
A= 1/3+1/3^2+1/3^3+1/3^4+1/3^5
3A=1+1/3+1/3^2+1/3^3+1/3^4
3A-A=1+1/3+1/3^2+1/3^3+1/3^4-1/3-1/3^2-1/3^3-1/3^4-1/3^5
2A=1-1/3^5
2A=242/243
A=121/243
A = 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/59049
3 x A = 3 x ( 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/59049 )
3 x A = 1 + 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/19683
3 x A - A = 1 + 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/19683
- ( 1 + 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/59049 )
= 1 - 1/59049
2 x A = 59048/59049
A = 59048/59049 : 2
A = 29524/59049