Tính tổng: 1 + 1/4+ 1/12 + 1/36 + 1/108 +1/324 + 1/972
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
M
Mai
12 tháng 11 2017
~1,37448559670782
hoặc
1 91/243
Đúng(0)
Những câu hỏi liên quan
NM
1
1 tháng 3 2019
Đặt tổng trên là A ta có
A/3=1/12+1/36+1/108+ +1/2916+1/8748
A-A/3=2A/3=
=1/4-1/8748=2186/8748
=>A=1093/1916
k mk nhá
Mk đã giúp bn rồi đấy
Ai k mk ,mk k lại,cảm ơn các bn
HC TỐT NHÁ BN
#HYISTHEBEST#
8 tháng 5 2018
Đặt A= 1/4+1/12+1/36+1/108+1/324 +1/972
=1/4.(1+1/3+1/9+1/27+1/81+1/243)
=1/4(1+1/3 +1/3^2 +1/3^3 +1/3^4 +1/3^5)
gọi M=1+1/3+1/3^2+1/3^3+1/3^4+1/3^5
3M=3+1+1/3+1/3^2+1/3^3+1/3^4
3M-M=(3+1+1/3+1/3^2+1/3^3+1/3^4)-(1+1/3+1/3^2+1/3^3+1/3^4+1/3^5)
2M=3-1/3^5
M=3/2-1/3^5.2
A=1/4.(3/2-1/3^5.2)
A=3/8-1/1944=91/243
Vậy .......
Chúc bạn học tốt .....^-^
3 tháng 4 2017
3xB=3x(1/4+1/12+1/36+1/108+1/324+1/972+1/2916+1/8748)
3xB=3/4 + 1/4 +1/12 +1/36 +.........+1/2916
3xB - B= (3/4 + 1/4 + 1/12+1/36 + .........+1/2916) - ( 1/4 +1/12 +1/36 +1/108 + 1/324 + 1/972 + 1/2916 +1/8748 )
2xB =3/4 - 1/8748
2xB =1640/2187
B = 1640/2187 :2
B = 820/2187.
21 tháng 7 2019
\(A=\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}\)
\(3A=3\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}+\frac{1}{972}\right)\)
\(3A=\frac{3}{4}+\frac{3}{12}+\frac{3}{36}+\frac{3}{108}+\frac{3}{324}+\frac{3}{927}\)
\(3A=\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}\)
\(2A=3A-A\)
\(2A=\left(\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}\right)-\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}+\frac{1}{972}\right)\)
\(2A=\frac{3}{4}-\frac{1}{927}\)
\(2A=\frac{729-1}{972}=\frac{728}{972}=\frac{182}{243}\)
\(A=\frac{182}{243}:\frac{1}{2}\)
\(A=\frac{364}{243}\)
NT
2
A
21 tháng 7 2019
\(A=\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}+\frac{1}{972}.\)
\(3A=3\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}+\frac{1}{972}\right).\)
\(3A=\frac{3}{4}+\frac{3}{12}+\frac{3}{36}+\frac{3}{108}+\frac{3}{324}+\frac{3}{972}\)
\(3A=\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}\)
\(2A=\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}-\frac{1}{4}-\frac{1}{12}-\frac{1}{36}-\frac{1}{108}-\frac{1}{324}-\frac{1}{972}\)
\(2A=\frac{3}{4}-\frac{1}{972}=\frac{182}{243}\)
\(\Rightarrow A=\frac{182}{243}:2=\frac{91}{243}\)
\(\Rightarrow A=\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}+\frac{1}{972}=\frac{91}{243}\)
21 tháng 7 2019
ta có
A x 3 =3/4 + 3/12 + 3/36 +3/108 + 3/324 +3/972
A x 3=3/4 +1/4 + 1/12 +1/36 +1/108
A x 3 - A =(3/4 +1/12+1/36 +1/108)-(1/4 +1/12 +1/36 +1/108 +1/324 + 1/972)
A x 2=3/4 +1/12 +1/36 +1/108 - 1/4 -1/12 -1/36-1/108 -1/324 -1/972
A x 2= 3/4 - 1/972
A x 2= 728/972
A =728/972 : 2
A=91/243
T
0
PA
1
26 tháng 3 2018
\(B=\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}+\dfrac{1}{972}\\\)
\(3B=3\left(\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}+\dfrac{1}{972}\right)\)
\(3B=\dfrac{3}{4}+\dfrac{3}{12}+\dfrac{3}{36}+\dfrac{3}{108}+\dfrac{3}{324}+\dfrac{3}{972}\)
\(3B=\dfrac{3}{4}+\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}\)
\(2B=3B-B\)
\(2B=\left(\dfrac{3}{4}+\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}\right)-\left(\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{36}+\dfrac{1}{108}+\dfrac{1}{324}+\dfrac{1}{972}\right)\)
\(2B=\dfrac{3}{4}-\dfrac{1}{972}=\dfrac{729-1}{972}=\dfrac{728}{972}=\dfrac{182}{243}\)
\(B=\dfrac{182}{243}:\dfrac{1}{2}=\dfrac{182\cdot2}{243}=\dfrac{364}{243}\)