K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
29 tháng 6 2020
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)
\(=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)
Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}+\frac{1}{2^9}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
\(\Rightarrow A=1-\frac{1}{2^{10}}=\frac{2^{10}-1}{2^{10}}\)
vậy \(A=\frac{2^{10}-1}{2^{10}}\)
DA
9 tháng 6 2019
Mk lm tắt nha
Hk bik có đúng hk nữa
A= 1/2-1/12024
.... Tự tính kết quả nha
9 tháng 6 2019
gọi A = 1/2+1/4+1/8+......+1/1024 2 x A = 1 + 1/2 + 1/4+......+1/1024 2 x A - A = (1+1/2+1/4 +....+1/512) - (1/2+1/4+1/8+.....1/1024) A= 1-1/1024 = 1023/1024 vậy A= 1023/1024
24 tháng 6 2016
A x 2 = 1 - ( 1/2 + 1/4 + 1/8 + 1/16 + ..... + 1/512 + 1/1024 ) - 1/1024
A x 2 = 1 - 1/1024 + A
A x 2 - A = 1 - 1/1024
A = 1 - 1/1024
A = 1023 /1024
24 tháng 6 2016
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
29 tháng 3 2018
A = 1 - 1/2 + 1/2 - 1/4 + 1/4 - 1/8 + ..... + 1/256 - 1/512 + 1/512 - 1/1024
A = 1 - 1/1024
A = 1023/1024
10 tháng 4 2020
Ta có: A=1/2 + 1/4 + 1/8 +...+ 1/512 + 1/1024 2A= 1+ 1/2 + 1/4+...+1/256+1/512 2A - A=(1+1/2+1/4+...+1/256+1/512) - ( 1/2+1/4+1/8+...+1/512+1/1024) suy ra A= 1 - 1/1024= 1023/1024 . Vậy A = 1023/1024. nhớ chọn câu trả lời của mình nha!
NH
10 tháng 4 2020
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{512}+\frac{1}{1024}\)
Đặt tổng trên là A , ta có :
\(A\times2=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\right)\)
\(A\times2=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\right)\times2\)
\(A\times2=\frac{1}{2}\times2+\frac{1}{4}\times2+\frac{1}{8}\times2+...+\frac{1}{512}\times2+\frac{1}{1024}\times2\)
\(A\times2=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)
\(A\times2-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\right)\)
\(A=\left(1-\frac{1}{1024}\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{1}{8}-\frac{1}{8}\right)+....+\left(\frac{1}{256}-\frac{1}{256}\right)+\left(\frac{1}{1024}-\frac{1}{1024}\right)\)
\(A=\frac{1023}{1024}\)
Nguyễn Linh Chi : Mong cô check bài này cho em ạ !!
NT
2
CS
6 tháng 10 2021
\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{512}+\dfrac{1}{1024}\)
\(=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)
\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)
\(\Rightarrow2A-A=A=1-\dfrac{1}{2^{10}}\)
DL
1
TG
17 tháng 2 2022
uyjbiyjhnbgyhgyuihyygbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb
=1-(1/2+1/2-1/4+1/4-1/8+1/8...-1/1024+1/1024-1/1024)
=1-1/1024
=1023/1024
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)
\(=1-\frac{1}{1024}\)
\(=\frac{1024}{1024}-\frac{1}{1024}\)
\(=\frac{1023}{1024}\)