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BT
4
NV
0
U
28 tháng 7 2018
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{256}-\frac{1}{512}+\frac{1}{512}-\frac{1}{1028}\)
\(=1-\frac{1}{1028}\)
\(=\frac{1027}{1028}\)
28 tháng 7 2018
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
\(A=1-\frac{1}{2^{10}}\)
\(A=\frac{2^{10}-1}{2^{10}}\)
Tham khảo nhé~
12 tháng 8 2016
A=\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{512}-\frac{1}{1024}\)
=1-1/1024
=1023/1024
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
6 tháng 4 2023
A = ( 1 - \(\dfrac{1}{2}\) ) + ( 1 - \(\dfrac{1}{4}\)) + ( 1 - \(\dfrac{1}{8}\)) +......+ ( 1 - \(\dfrac{1}{512}\)) + ( 1 - \(\dfrac{1}{1024}\))
A = (1 + 1 +....+ 1) - ( \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + ......+ \(\dfrac{1}{512}\) + \(\dfrac{1}{1024}\))
A = ( 1 + 1 +.....+ 1) - ( \(\dfrac{1}{2^1}\) + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\)+....+ \(\dfrac{1}{2^9}\) + \(\dfrac{1}{2^{10}}\))
Vì trong tổng A có 10 phân số nên
nhóm ( 1 + 1 +....+ 1) có 10 hạng tử là 1
Vậy A = 1 \(\times\) 10 - ( \(\dfrac{1}{2^1}\) + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\) +..........+ \(\dfrac{1}{2^9}\) + \(\dfrac{1}{2^{10}}\))
Đặt B = \(\dfrac{1}{2^1}\) + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\) +......+ \(\dfrac{1}{2^9}\) + \(\dfrac{1}{2^{10}}\)
2 \(\times\) B = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\)+........+ \(\dfrac{1}{2^9}\)
2B - B = 1 - \(\dfrac{1}{2^{10}}\)
B = 1 - \(\dfrac{1}{2^{10}}\)
A = 10 + 1 - \(\dfrac{1}{2^{10}}\)
A = 11 - \(\dfrac{1}{2^{10}}\)
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
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)
\(2A-A=\left[1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\right]-\left[\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\right]\)
\(A=1-\frac{1}{2014}=\frac{2013}{2014}\)
15 tháng 8 2018
\(A=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^9}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
\(A=1-\frac{1}{2^{10}}\)
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
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{10}}\)
\(\Leftrightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^9}\)
\(\Rightarrow2A-A=1-\frac{1}{2^{10}}=\frac{1023}{1024}\)