hãy tính: M=1/3+1/9+1/27+1/81+.....+1/6561
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Những câu hỏi liên quan
1 tháng 7 2018
M = \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{6561}\)
=> 3M = \(1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{2187}\)
=> 3M - M = ( \(1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{2187}\) ) - ( \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{6561}\))
2M = 1 - \(\frac{1}{6561}\)
2M = \(\frac{6560}{6561}\)
=> M = \(\frac{3280}{6561}\)
HN
1 tháng 7 2018
\(M=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+.......+\frac{1}{6561}\)
\(\Rightarrow M=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+.........+\frac{1}{3^8}\)
\(\Rightarrow3M=3\left(\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+.........+\frac{1}{3^8}\right)\)
\(\Rightarrow3M=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+............+\frac{1}{3^7}\)
\(\Rightarrow3M-M=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+..........+\frac{1}{3^7}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-.......-\frac{1}{3^8}\)
\(\Rightarrow2M=1-\frac{1}{3^8}\)
\(\Rightarrow M=\frac{1-\frac{1}{3^8}}{2}\)
Vậy M = \(\frac{1-\frac{1}{3^8}}{2}\)
CL
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
DT
1
8 tháng 5 2015
Cho \(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{6561}\)
\(\frac{1}{3}A=\frac{1}{3}\times\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{6561}\right)\)
\(\frac{1}{3}A=\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+...+\frac{1}{19683}\)
\(A-\frac{1}{3}A=\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{6561}\right)-\left(\frac{1}{9}+\frac{1}{27}+...+\frac{1}{19683}\right)\)
\(\frac{2}{3}A=\frac{1}{3}-\frac{1}{19683}\)
\(A=\frac{4840}{9683}:\frac{2}{3}=\frac{7260}{9683}\)
TN
14 tháng 8 2017
\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+...+\frac{1}{6561}\)
\(\Rightarrow A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^8}\)
\(\Rightarrow3A=3.\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)\) \(=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)
\(\Rightarrow3A-A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-...-\frac{1}{3^8}\)
\(\Rightarrow2A=1-\frac{1}{3^8}\) \(\Rightarrow A=\frac{1-\frac{1}{3^8}}{2}\)
k cho mik đi mn!Nguyễn Như Quỳnh!
ES
1
ND
Nguyễn Đức Trí
VIP
14 tháng 8 2023
\(A=\dfrac{3}{5.6}+\dfrac{3}{6.7}+...+\dfrac{3}{91.92}\)
\(\Rightarrow A=3\left(\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{91.92}\right)\)
\(\Rightarrow A=3\left(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{91}-\dfrac{1}{92}\right)\)
\(\Rightarrow A=3\left(\dfrac{1}{5}-\dfrac{1}{92}\right)\)
\(\Rightarrow A=3.\dfrac{87}{460}=\dfrac{261}{460}\)
TD
1
14 tháng 3 2023
\(A=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+...+\dfrac{1}{2187}+\dfrac{1}{6561}\)
\(3A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+...+\dfrac{1}{2187}\)
Lấy 3A - A ta được :
\(2A=1-\dfrac{1}{6561}=\dfrac{6560}{6561}\Leftrightarrow A=\dfrac{6560}{6561}:2\)
\(\Leftrightarrow A=\dfrac{6560}{6561}.\dfrac{1}{2}=\dfrac{3280}{6561}\)
HD
1
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 : ...
OD
3
H9
HT.Phong (9A5)
CTVHS
9 tháng 7 2023
\(B=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+...+\dfrac{1}{2187}+\dfrac{1}{6561}\)
\(3B=3\cdot\left(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+...+\dfrac{1}{6561}\right)\)
\(3B=1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{729}+\dfrac{1}{2187}\)
\(3B-B=\left(1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{2187}\right)-\left(\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{6561}\right)\)
\(2B=\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{9}-\dfrac{1}{9}\right)+...+\left(1-\dfrac{1}{6561}\right)\)
\(2B=0+0+...+1-\dfrac{1}{6561}\)
\(2B=1-\dfrac{1}{6561}\)
\(B=\left(1-\dfrac{1}{6561}\right):2\)
\(B=\dfrac{6560}{6561}:2\)
\(B=\dfrac{3280}{6561}\)
3M=1+1/3+1/9+...+1/2187
2M=3M-M
2M=1-1/6561
2M=6560/6561
M=3280/6561