\(\frac{1}{3}\)+\(\frac{1}{9}\)+\(\frac{1}{27}\)+\(\frac{1}{61}\)+.......+\(\frac{1}{2187}\)=
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Những câu hỏi liên quan
NA
4
13 tháng 5 2017
A=1/3+1/9+1/27+...+1/2187
=1/3+1/3^2+1/3^3+...+1/3^7
-->3A=1+1/3+1/3^2+...+1/3^6
-->3A-A=(1+1/3+1/3^2+...+1/3^6) - (1/3+1/3^2+1/3^3+...+1/3^7)
-->2A=1- 1/3^7
-->A=1093/2187
NL
1
T
14 tháng 5 2019
#)Giải :
\(A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)
\(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}\)
\(A=\frac{2187}{2187}+\frac{729}{2187}+\frac{243}{2187}+\frac{81}{2187}+\frac{27}{2187}+\frac{9}{2187}+\frac{3}{2187}+\frac{1}{2187}\)
\(A=\frac{3037}{2187}\)
#~Will~be~Pens~#
NN
0
NN
0
PT
4
KT
26 tháng 7 2018
\(S=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\)
\(=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)
\(\Rightarrow\)\(3S=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\)
\(\Rightarrow\)\(3S-S=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)\)
\(\Rightarrow\)\(2S=1-\frac{1}{3^7}\)
\(\Rightarrow\)\(S=\frac{1-\frac{1}{3^7}}{2}\)
26 tháng 7 2018
\(S=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)
\(3S=1+\frac{1}{3}+...+\frac{1}{3^6}\)
\(3S-S=\left(1+\frac{1}{3}+...+\frac{1}{3^6}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)\)
\(2S=1-\frac{1}{3^7}\)
\(S=\frac{1-\frac{1}{3^7}}{2}\)
12 tháng 6 2015
Gọi tong trên là A
\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{81}+\frac{1}{243}+\frac{1}{7129}+\frac{1}{2187}\)
\(3A=\frac{1}{3}+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{729}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\right)-\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}\right)\)
\(2A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}-\frac{1}{3}-\frac{1}{9}-\frac{1}{27}-\frac{1}{81}-\frac{1}{243}-\frac{1}{729}-\frac{1}{2187}\)
\(2A=1-\frac{1}{2187}\)
\(2A=\frac{2186}{2187}\)
\(A=\frac{2186}{2187}:2\)
\(A=\frac{1093}{2187}\)
Vậy tổng A = \(\frac{1093}{2187}\)
12 tháng 6 2015
\(3y=3\cdot\frac{1}{1}+3\cdot\frac{1}{3}+3\cdot\frac{1}{9}+...+3\cdot\frac{1}{729}+3\cdot\frac{1}{2187}\)
\(=3+\frac{1}{1}+\frac{1}{3}...+\frac{1}{729}\)
=> \(3y-y=3+\frac{1}{1}+\frac{1}{3}+..+\frac{1}{729}-\frac{1}{1}-\frac{1}{3}-...-\frac{1}{2187}\)
<=> 2y = 3- 1/2187
=> y = \(\frac{3-\frac{1}{2187}}{2}\)
BN
2
30 tháng 1 2015
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)
\(\Rightarrow\frac{1}{3}B=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^8}\)
\(\Rightarrow B-\frac{1}{3}B=\frac{1}{3}-\frac{1}{3^8}\Rightarrow\frac{2}{3}B=\frac{3^7-1}{3^8}\Rightarrow B=\frac{3\left(3^7-1\right)}{2.3^8}\)
4 tháng 3 2018
Ta có :
\(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\)
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\)
\(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\right)\)
\(2B=1-\frac{1}{3^7}\)
\(2B=\frac{3^7-1}{3^7}\)
\(B=\frac{3^7-1}{3^7}:2\)
\(B=\frac{3^7-1}{2.3^7}\)
Vậy \(B=\frac{3^7-1}{2.3^7}\)
Chúc bạn học tốt ~
16 tháng 8 2017
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
\(=[3\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{81}+...+\frac{1}{2187}\right)-\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{81}+...+\frac{1}{2187}\right)]:2\)
\(=\left(1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{729}-\frac{1}{3}-\frac{1}{9}-\frac{1}{81}-...-\frac{1}{2187}\right):2\)
\(=\left(1-\frac{1}{2187}\right):2=\frac{2186}{2187}.\frac{1}{2}=\frac{1093}{2187}\)