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NV
2
14 tháng 8 2023
Sửa đề: A=1/3+1/9+1/27+...+1/6561
=1/3+1/3^2+1/3^3+...+1/3^8
=>3A=1+1/3+...+1/3^7
=>3A-A=1-1/3^8
=>\(2A=\dfrac{3^8-1}{3^8}\)
=>\(A=\dfrac{3^8-1}{2\cdot3^8}\)
NN
14 tháng 8 2023
Đặt \(S=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{6561}\)
\(3S=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{2187}\)
\(2S=\dfrac{2188}{2187}-\left(\dfrac{1}{27}+\dfrac{1}{6561}\right)\)
\(2S=\dfrac{2188}{2187}-\dfrac{244}{6561}\)
\(2S=\dfrac{4376}{6561}-\dfrac{244}{6561}\)
\(2S=\dfrac{4132}{6561}\)
\(S=\dfrac{2066}{6561}\)
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
22 tháng 8 2018
\(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{6561}\)
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)
\(3B-B=\left(1+\frac{1}{3}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)\)
\(2B=1-\frac{1}{3^8}\)
\(B=\frac{1-\frac{1}{3^8}}{2}\)
SF
22 tháng 8 2018
B = 1/3 + 1/9 + 1/27 + ... + 1/6561
B = 1/3^1 + 1/3^2 + 1/3^3 + ... + 1/3^8
3B = 1 + 1/3^1 + 1/3^2 + ... + 1/3^7
3B - B = ( 1 + 1/3^1 +1/3^2 + ... + 1/3^7 ) - ( 1/3^1 + 1/3^2 + 1/3^3 + .... + 1/3^8 )
2B = 1 - 1/3^8
B = 1 - 1/3^8 / 2
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
25 tháng 7 2023
(y + \(\dfrac{1}{3}\)) + ( y + \(\dfrac{1}{9}\)) + ( y + \(\dfrac{1}{27}\)) + ( y + \(\dfrac{1}{81}\)) = \(\dfrac{56}{81}\)
( y + y + y + y ) + (\(\dfrac{1}{3}\)+ \(\dfrac{1}{9}\) + \(\dfrac{1}{27}\) + \(\dfrac{1}{81}\)) = \(\dfrac{56}{81}\)
4\(y\) + ( \(\dfrac{27}{81}\) + \(\dfrac{9}{81}\) + \(\dfrac{3}{27}\) + \(\dfrac{1}{81}\) ) = \(\dfrac{56}{81}\)
4y + \(\dfrac{40}{81}\) = \(\dfrac{56}{81}\)
4y = \(\dfrac{56}{81}\) - \(\dfrac{40}{81}\)
4y = \(\dfrac{16}{81}\)
y = \(\dfrac{16}{81}\) : 4
y = \(\dfrac{4}{81}\)
H9
HT.Phong (9A5)
CTVHS
25 tháng 7 2023
\(\left(y+\dfrac{1}{3}\right)+\left(y+\dfrac{1}{9}\right)+\left(y+\dfrac{1}{27}\right)+\left(y+\dfrac{1}{81}\right)=\dfrac{56}{81}\)
\(\Rightarrow y+\dfrac{1}{3}+y+\dfrac{1}{9}+y+\dfrac{1}{27}+y+\dfrac{1}{81}=\dfrac{56}{81}\)
\(\Rightarrow4\times y+\dfrac{40}{81}=\dfrac{56}{81}\)
\(\Rightarrow4\times y=\dfrac{56}{81}-\dfrac{40}{81}\)
\(\Rightarrow4\times y=\dfrac{16}{81}\)
\(\Rightarrow y=\dfrac{16}{81}:4\)
\(\Rightarrow y=\dfrac{4}{81}\)
LS
10 tháng 7 2018
T = 5.5 + 6.6 + .... + 30.30
T = 5.(6 - 1) + 6.(7-1) + ... + 30.(31 - 1)
T = 5.6 - 5 + 6.7 - 6 + ... + 30.31 - 30
T = (5.6 + 6.7 + ... + 30.31) - (5 + 6 + ... + 30)
Đặt A = 5.6+ 6.7 + ... + 30.31
B = 5 + 6 + ... + 30
Ta có :
3A = 5.6.3 + 6.7.3 + ... + 30.31 . 3
3A = 5.6.(7-4) + 6.7.(8-5) + ... + 30.31.(32-29)
3A = 5.6.7 - 4.5.6 + 6.7.8 - 5.6.7 + ... + 30.31.32 - 29.30.31
3A = (5.6.7 + 6.7.8 + ... + 30.31.32) - (4.5.6 + 5.6.7 + ... + 29.30.31)
3A = 30.31.32 - 4.5.6
3A = 29640
A = 29640 : 3
A = 9880
SSH của B là : (30 - 5) : 1 + 1 = 26 (số hạng)
Tổng B là : (30 + 5) . 26 : 2 =455
=> T = A - B = 9880 - 455 = 9425
c, S = 1 + 3 + 9 + 27 + 81 + 243 + 729 + 2187 + 6561
S = (3 + 2187) + (9 + 6561) + (27 + 243) + (81 + 729) + 1
S = 2190 + 6570 + 270 + 810 + 1
S = (2190 + 810) + 6570 + 270 + 1
S = 3000 + 6570 + 270 + 1
S = 9570 + 270 + 1
S = 9840 + 1
S = 9841
Vậy S = 9841
TN
6 tháng 7 2016
Đặt A=1/3 + 1/9 + 1/27 + 1/81 + 1/24 + 1/729
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^6}\)
\(3A=3\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\right)\)
\(3A=1+\frac{1}{3}+...+\frac{1}{3^5}\)
\(3A-A=\left(1+\frac{1}{3}+...+\frac{1}{3^5}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\right)\)
\(2A=1-\frac{1}{3^6}\)
\(A=\frac{1-\frac{1}{3^6}}{2}\)
5 tháng 8 2016
\(\text{Đ}\text{ặt}:A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(\Rightarrow3A=3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(\Rightarrow3A-A=3-\frac{1}{729}\)
\(\Rightarrow2A=\frac{2186}{729}\)
\(\Rightarrow A=\frac{2186}{729}:2=\frac{1093}{729}\)
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 = \(\frac{3280}{6561}\)
\(M=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+.......+\frac{1}{6561}\)
\(\Rightarrow M=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+.........+\frac{1}{3^8}\)
\(\Rightarrow3M=3\left(\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+.........+\frac{1}{3^8}\right)\)
\(\Rightarrow3M=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+............+\frac{1}{3^7}\)
\(\Rightarrow3M-M=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+..........+\frac{1}{3^7}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-.......-\frac{1}{3^8}\)
\(\Rightarrow2M=1-\frac{1}{3^8}\)
\(\Rightarrow M=\frac{1-\frac{1}{3^8}}{2}\)
Vậy M = \(\frac{1-\frac{1}{3^8}}{2}\)