tính nhanh:
\(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{512}+\frac{1}{1024}\)
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TT
3
XO
2 tháng 2 2020
Ta có : \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)
Đặ A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)(1)
=> 2A = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}+\frac{1}{2^9}\)(2)
Lấy (2) trừ (1) theo vế ta có :
2A - A = \(\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\right)\)
=> A = \(1-\frac{1}{2^{10}}=\frac{2^{10}-1}{2^{20}}\)
2 tháng 2 2020
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{10}}\)
\(\Leftrightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^9}\)
\(\Rightarrow2A-A=1-\frac{1}{2^{10}}=\frac{1023}{1024}\)
10 tháng 5 2017
\(2A=2\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^9}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
\(A=1-\frac{1}{2^{10}}=\frac{2^{10}-1}{2^{10}}=\frac{1023}{1024}\)
BẤM ĐÚNG NHÉ
S
19 tháng 6 2019
\(2A=1+\frac{1}{2}+\frac{1}{4}+....+\frac{1}{512}\Rightarrow2A-A=1-\frac{1}{1024}=\frac{1023}{1024}\)
19 tháng 6 2019
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)
\(2A-A=\left[1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\right]-\left[\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\right]\)
\(A=1-\frac{1}{2014}=\frac{2013}{2014}\)
DA
tính biểu thức sau
\(a=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+..........+\frac{1}{512}+\frac{1}{1024}\)
2
29 tháng 7 2017
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)
\(2A=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{10}}+\frac{1}{2^{11}}\)
\(2A-A=\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{10}}+\frac{1}{2^{11}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\right)\)
\(A=2^{11}-2\)
PL
10 tháng 6 2018
(1981 x 1982 - 990) : (1980 x 1982 + 992)
=(1980 x 1982+1982 -990) : (1980 x 1982 +992)
=(1980 x 1982 + 992) : ( 1980 x 1982 + 992)
=1
TT
6
30 tháng 3 2018
Ta có :
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
\(A=1-\frac{1}{2^{10}}\)
\(A=\frac{2^{10}-1}{2^{10}}\)
\(A=\frac{1024-1}{1024}\)
\(A=\frac{1023}{1024}\)
Vậy \(A=\frac{1023}{1024}\)
Chúc bạn học tốt ~
30 tháng 3 2018
Đặt tổng trên là A.
Ta có
A x 2 = 1+ 1/2+1/4+1/8+ 1/16+1/32+ 1/64+ 1/128 + 1/256 + 1/512
Ax2 - A = 1+ 1/2+1/4+1/8 +1/16 + 1/32 +1/64+ 1/128 + 1/256+ 1/512 - ( 1/2 + 1/4 +1/8+1/16+1/32+1/64 + 1/128+ 1/256 + 1/512+ 1/1024)
A = 1+ 1/2 +1/4+1/8+1/16+1/32+1/64+1/128+1/256 + 1/512 - 1/2-1/4-1/8-1/16-1/32-1/64-1/128-1/256-1/512- 1/1024
A = 1 - 1/ 1024 = 1023/1024
IL
2 tháng 4 2017
ta có: 1+1/2+2+1/4+...+9+1/512
=(1+2+3+4+...+9)+(1/2+1/4+...+1/512)
=45+(1/2+1/4+...+1/512)
gọi số hạng (1/2+1/4+...+1/512) là a ta được :
a=1/2+1/4+...+1/512
2a=1+1/2+1/4+1/8+...+1/256
2a-a=(1+1/2+1/4+...+1/256)-(1/2+1/4+...+1/512)
=1-1/512
=511/512
vậy kết quả của biểu thức đó là45+511/512
M
15 tháng 8 2017
ta có : A=1/2+1/4+..+1/1024
=> A=1/21+1/22+..+1/210
=> A.2=(1/21+1/22+..+1/210).2
=> A.2=1+1/21+1/22+..+1/29
=> 2A-A=(1+1/21+1/22+..+1/29)-(1/21+1/22+..+1/210)
=> A=1-1/210
NY
2
U
15 tháng 8 2018
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{512}-\frac{1}{1024}\)
\(=1-\frac{1}{1024}\)
\(=\frac{1023}{1024}\)
M
15 tháng 8 2018
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}.\)
Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)
<=> \(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}+\frac{1}{512}\)
<=> \(2A-A=1+\frac{1}{2}+\frac{1}{4}+....+\frac{1}{256}+\frac{1}{512}-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{512}-\frac{1}{1024}\)
<=> \(A=1-\frac{1}{1024}\)
<=> \(A=\frac{1023}{1024}\)
A
21 tháng 6 2018
\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+....+\frac{1}{256}-\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{512}\)
\(=\frac{255}{512}\)
Vậy \(A=\frac{255}{512}\)
BD
21 tháng 6 2018
A=14 +18 +116 +132 +164 +1128 +1256 +1512
=12 −14 +14 −18 +....+1256 −1512
=12 −1512
=255512
Vậy A=255512
Phạm Long Khánh
Ta có: \(\frac{1}{2}=1-\frac{1}{2}\); \(\frac{1}{4}=\frac{1}{2}-\frac{1}{4}\); \(\frac{1}{8}=\frac{1}{4}-\frac{1}{8}\); ...; \(\frac{1}{512}=\frac{1}{256}-\frac{1}{512}\); \(\frac{1}{1024}=\frac{1}{512}-\frac{1}{1024}\)
Vậy \(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{256}-\frac{1}{512}+\frac{1}{512}-\frac{1}{1024}\)
\(=1+1-\frac{1}{1024}\)
\(=2-\frac{1}{1024}=\frac{2047}{1024}\)
bằng 2047/1024