Tính A=2/3+2/9+2/27+...+2/2187+2/6561 .Tính nhe mọi người, mình tick cho
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PH
2
NC
27 tháng 8 2020
Bài làm:
Ta có: \(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}+\frac{1}{6561}=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}+\frac{1}{3^8}\)
=> \(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}+\frac{1}{3^7}\)
=> \(3A-A=\left(1+\frac{1}{3}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)\)
<=> \(2A=1-\frac{1}{3^8}=\frac{3^8-1}{3^8}\)
=> \(A=\frac{3^8-1}{3^8.2}\)
B
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 !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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 : ...
T
3
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
TD
1
9 tháng 8 2019
Ta có:
\(S=3.2^0-3^1+3.2^1-3^2+3.2^2+3.2^3-3^3+3.2^4-3^4+...-3^7+3.2^{10}+3.2^{11}-3^8+3.2^{12}\)
\(=3.\left(2^0+2^1+2^2+2^3+2^4+...+2^{10}+2^{11}+2^{12}\right)-\left(3^1+3^2+3^3+...+3^7+3^8\right)\)
Đặt: \(A=2^0+2^1+2^2+...+2^{11}+2^{12}\)
=> \(2.A=2^1+2^2+2^3+...+2^{12}+2^{13}\)
=> \(2.A-A=2^{13}-2^0\)
\(\Rightarrow A=2^{13}-1=8191\)
Đặt: \(B=3^1+3^2+3^3+...+3^8\)
\(\Rightarrow3.B=3^2+3^3+3^4+...+3^9\)
=> \(3B-B=3^9-3^1=19680\)
=> \(2B=19680\Rightarrow B=9840\)
=> S=3.A-B=3.8191-9840=14733
TN
1
24 tháng 6 2023
\(3A=1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{2187}\)
\(3A-A=\left(1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{2187}\right)-\left(\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{6561}\right)\)
\(2A=\dfrac{6560}{6561}\)
\(A=\dfrac{3280}{6561}\)
14 tháng 8 2016
đặt A = 1+3+9+27+....+2187+6561
=>A = 30 + 31 + 32 + 33 + .. . +37 + 38
3A = 31 + 32 + 33 + ... + 38 + 39
3A - A = (31 + 32 + 33 + ... + 38 + 39)-(30 + 31 + 32 + 33 + .. . +37 + 38 )
2A = 39 - 1
A=\(\frac{3^9-1}{2}=\frac{19682}{2}=9841\)
\(A=\frac{2}{3}+\frac{2}{9}+\frac{2}{27}+...+\frac{2}{2187}+\frac{2}{6561}\)
\(3\times A=2+\frac{2}{3}+\frac{2}{9}+...+\frac{2}{729}+\frac{2}{2187}\)
\(3\times A-A=\left(2+\frac{2}{3}+\frac{2}{9}+...+\frac{2}{2187}\right)-\left(\frac{2}{3}+\frac{2}{9}+\frac{2}{27}+...+\frac{2}{2187}+\frac{2}{6561}\right)\)
\(2\times A=2-\frac{2}{6561}\)
\(A=\frac{6560}{6561}\)