A=1+1/3+1/9+1/27+...+1/729x3
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NT
1
AH
Akai Haruma
Giáo viên
28 tháng 9
Lời giải:
Đặt $A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}$
$3\times A=3+1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{729}$
$3\times A-A=3-\frac{1}{2187}$
$2\times A=3-\frac{1}{2187}=\frac{6560}{2187}$
$A=\frac{6560}{2187}:2=\frac{3280}{2187}$
NT
2
20 tháng 4 2017
\(=1+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)
\(=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}\)
\(=\frac{2187}{2187}+\frac{729}{2187}+\frac{81}{2187}+\frac{27}{2187}+\frac{9}{2187}+\frac{3}{2187}+\frac{1}{2187}\)
\(=\frac{2187+729+81+27+9+3+1}{2187}\)
\(=\frac{3037}{2187}\)
Đúng 100%
PT
2
DL
27 tháng 2 2018
\(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{729\cdot3}\)
\(A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)
\(3A=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\)
\(3A-A=\left(4+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)\)
\(2A=3-\frac{1}{3^7}\)
\(A=\frac{ 1}{2}\left(3-\frac{1}{3^7}\right)\)
VL
1
BT
5 tháng 12 2017
\(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{729.3}\)
\(A=1+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)
=> \(3A=3+1+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^6}\)
=> \(3A-A=3+1+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^6}-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)\)
<=> \(2A=3-\frac{1}{3^7}=\frac{3^8-1}{3^7}\)
=> \(A=\frac{3^8-1}{2.3^7}\)
MT
0
TN
1
24 tháng 6 2023
\(3A=1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{2187}\)
\(3A-A=\left(1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{2187}\right)-\left(\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{6561}\right)\)
\(2A=\dfrac{6560}{6561}\)
\(A=\dfrac{3280}{6561}\)
17 tháng 8 2023
1:
a: =23/27-11/17+4/27+28/17
=23/27+4/27+28/17-11/17
=1+1=2
b: \(=\dfrac{2}{3}\cdot\left(\dfrac{7}{9}+\dfrac{2}{9}\right)-\dfrac{2}{9}\)
=2/3-2/9
=6/9-2/9
=4/9
c: \(=\dfrac{11}{5}\cdot\dfrac{7}{3}-\dfrac{1}{3}\cdot\dfrac{11}{5}\)
=11/5(7/3-1/3)
=11/5*2
=22/5
d: \(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{2024}{2023}=\dfrac{2024}{2}=1012\)
e: \(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{2022}{2023}=\dfrac{1}{2023}\)
Dấu "." là dấu nhân bạn nhé.
Ta có:
\(A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+...+\dfrac{1}{729.3}\)
\(\Rightarrow3A=3+1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{729}\)
\(\Rightarrow3A-A=3-\dfrac{1}{729.3}\)
\(\Rightarrow2A=3-\dfrac{1}{729.3}\)
\(\Rightarrow A=\dfrac{1}{2}\left(3-\dfrac{1}{729.3}\right)\)