Tính: \(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\). Cần gấp!
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
PA
13 tháng 9 2016
\(A=\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(2A=\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-..-\frac{1}{512}\)
\(2A-A=\left(\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-..-\frac{1}{512}\right)-\left(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\right)\)
\(A=\frac{1}{4}+\frac{1}{4}-\frac{1}{2}+\frac{1}{1024}\)
\(A=\frac{1}{1024}\)
13 tháng 9 2016
\(B=\frac{1}{2}-\frac{1}{4}-...-\frac{1}{1024}\)
\(=-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)\)
\(=-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
Đặt \(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}=A\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^9}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
\(A=1-\frac{1}{2^{10}}\).Thay A vào ta đc: \(B=-\left(1-\frac{1}{2^{10}}\right)\)
\(B=-\left(1-\frac{1}{1024}\right)\)
\(B=-\frac{1023}{1024}\)
PA
5 tháng 9 2016
\(A=-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(A=-\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
\(-A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(-2A=2+1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)
\(-2A+A=\left(2+1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
\(-A=2-\frac{1}{1024}\)
\(A=\frac{1}{1024}-2\)
BT
3
18 tháng 10 2016
\(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-....-\frac{1}{1024}\)
\(=1-\left(1-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{4}\right)-.......-\left(\frac{1}{512}-\frac{1}{1024}\right)\)
\(=1-1+\frac{1}{2}-\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+....+\frac{1}{512}-\frac{1}{1024}\)
\(=-\frac{1}{1024}\)
HT
18 tháng 10 2016
\(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-....-\frac{1}{1024}=1-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
=> \(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
=> \(A=2A-A=1-\frac{1}{2^{10}}\)
=> \(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-....-\frac{1}{1024}=1-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
\(=1-A=1-\left(1-\frac{1}{2^{10}}\right)=1-1+\frac{1}{2^{10}}\)
\(=\frac{1}{2^{10}}\)
SH
0
CV
0
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
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)
\(2A-A=\left[1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\right]-\left[\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\right]\)
\(A=1-\frac{1}{2014}=\frac{2013}{2014}\)
LL
3
11 tháng 9 2016
\(-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-.....-\frac{1}{1024}\)
\(=-1-\left(1-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{4}\right)-.....-\left(\frac{1}{512}-\frac{1}{1024}\right)\)
\(=-1-\left(1-\frac{1}{1024}\right)\)
\(=-1-\frac{1023}{1024}\)
\(=-\frac{2047}{1024}\)
11 tháng 9 2016
Ai mún nhờ mk giải violympic vòng 1 300 đ ko
10 nghìn 1 lần nhé hoặc là xóa nick facebook (20 nghìn 1 lần)
Hoặc Hack facebook (10 nghìn 1 lên 500 )
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
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{10}}\)
\(\Leftrightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^9}\)
\(\Rightarrow2A-A=1-\frac{1}{2^{10}}=\frac{1023}{1024}\)
TH
1
25 tháng 9 2016
\(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}=\frac{1}{1024}\)dùng phương pháp loại trừ
Đặt :
\(A=1+\frac{1}{2}+\frac{1}{4}+..........+\frac{1}{1024}\)
\(\Leftrightarrow A=1+\frac{1}{2}+\frac{1}{2^2}+..........+\frac{1}{2^{10}}\)
\(\Leftrightarrow2A=2+1+\frac{1}{2}+......+\frac{1}{2^9}\)
\(\Leftrightarrow2A-A=\left(2+1+\frac{1}{2}+....+\frac{1}{2^9}\right)-\left(1+\frac{1}{2}+....+\frac{1}{2^{10}}\right)\)
\(\Leftrightarrow A=2-\frac{1}{2^{10}}\)
Gọi dãy số trên là A
\(A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\)
\(2A=2+1+\frac{1}{2}+...+\frac{1}{512}\)
\(2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{512}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}+\frac{1}{1024}\right)\)
\(A=2-\frac{1}{1024}\)
\(A=\frac{2048}{1024}-\frac{1}{1024}\)
\(A=\frac{2047}{1024}\)