NP
2
K
Khách
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2 tháng 9 2021
\(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}\)
\(=3.\left(\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^9}\right)\)
\(=3.A\)với \(A=\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^9}\)
\(\Rightarrow2^2A=\left(2+\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}\right)\)
\(\Rightarrow2^2A-A=\left(2+\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}\right)-\left(\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^9}\right)\)
\(\Rightarrow4A-A=2-\frac{1}{2^9}\)
\(\Rightarrow3A=2-\frac{1}{512}=\frac{1023}{512}\Rightarrow A=\frac{1023}{512}:3\)
\(\Rightarrow\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}=3.\left(\frac{1023}{512}:3\right)=\frac{1023}{512}\)
NM
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
24 tháng 6 2023
C = \(\dfrac{3}{2}\) + \(\dfrac{3}{4}\) + \(\dfrac{3}{8}\)+ \(\dfrac{3}{16}\)+...........+\(\dfrac{3}{128}\)
C\(\times\)2 = 3 + \(\dfrac{3}{2}\) + \(\dfrac{3}{4}\) + \(\dfrac{3}{8}\) + \(\dfrac{3}{16}\)+...+\(\dfrac{1}{64}\)
C\(\times\)2 - C = 3 - \(\dfrac{3}{128}\)
C = \(\dfrac{381}{128}\)
Câu E em xem lại đề nhé
NM
2
SX
1
17 tháng 3 2016
3(1/2+1/4+1/8+1/16.....1/128)
A=(1/2+1/4+1/8+1/16....1/128)
2A-A=A
2A=(1/2+1/4+1/8....1/128)*2
=1+1/2+1/4....1/64
2A-A=1+1/2+1/4+1/8....1/64-(1/2+1/4+1/8......1/128)
A=1-1/128=127/128
Đáp số:3*127/128=381/128
KT
14 tháng 8 2018
a) \(D=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+...+\frac{1}{512}+\frac{1}{1024}\)
=> \(2D=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...++\frac{1}{256}+\frac{1}{512}\)
=> \(2D-D=\left(1+\frac{1}{2}+...+\frac{1}{512}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)\)
=> \(D=1-\frac{1}{1024}\)
b) \(Đ=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\)
\(=1-\frac{1}{20}=\frac{19}{20}\)
CM
14 tháng 8 2018
a) D=\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\dots+\frac{1}{512}+\frac{1}{1024}.\)
\(D=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\dots+\frac{1}{512}-\frac{1}{1024}\)
\(D=1-\frac{1}{1024}\)
\(D=\frac{1023}{1024}\)
\(Đ=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\dots+\frac{1}{18\cdot19}+\frac{1}{19\cdot20}\)
\(Đ=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\dots+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\)
\(Đ=1-\frac{1}{20}\)
\(Đ=\frac{19}{20}\)
Phần c như kiểu sai đề chỗ cuối hay sao ấy.
15 tháng 8 2016
\(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}\)
\(=\frac{192}{128}+\frac{48}{128}+\frac{12}{128}+\frac{3}{128}\)
\(=\frac{192+48+12+3}{128}=\frac{255}{128}\)
Đặt tổng trên là: \(A\)
\(A=\dfrac{3}{2}+\dfrac{3}{8}+\dfrac{3}{32}+\dfrac{3}{128}+\dfrac{3}{512}\)
\(\Rightarrow A.4=6+\dfrac{3}{2}+\dfrac{3}{8}+\dfrac{3}{32}+\dfrac{3}{128}\)
\(\Rightarrow A.4-A=\left(6+\dfrac{3}{2}+\dfrac{3}{8}+\dfrac{3}{32}+\dfrac{3}{128}\right)-\left(\dfrac{3}{2}+\dfrac{3}{8}+\dfrac{3}{32}+\dfrac{3}{128}+\dfrac{3}{512}\right)\)
\(\Rightarrow A.3=6-\dfrac{3}{512}=\dfrac{3069}{512}\)
\(\Rightarrow A=\dfrac{3069}{512}:3=\dfrac{1023}{512}\)
3/512 là gì vậy bn?