tìm tổng 1/2 + 1/4 + 1/8 +1/16+ 1/32 + 1/64 help mình huhuhuhuhuh
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
22 tháng 10 2016
S=1/2+1/4+1/8+1/16+1/32+1/64
S=1-1/2+1/2-1/4+1/4-1/8+1/8-1/16+1/16-1/32+1/32-1/64
S=1-1/64
S=63/64
PT
1
CM
24 tháng 6 2017
Cộng thêm 1/2 vào biểu thức đã cho, có:
S + 1/2= 1/2+1/4+ 1/8+ 1/16+1/32+1/64+1/128
Nhận xét:
QT
1
12 tháng 9 2018
Ta có : A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64
=> 2A = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32
=> 2A - A = 1 - 1/64
=> A = 1 - 1/64
=> A = 63/64
LL
4
5 tháng 4 2017
Ta có : \(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{64}\)
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^6}\)
\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^5}\)
\(\Rightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^5}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^6}\right)\)
\(\Rightarrow A=1-\dfrac{1}{2^6}=1-\dfrac{1}{64}=\dfrac{63}{64}\)
5 tháng 4 2017
\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}+...+\dfrac{1}{32}-\dfrac{1}{64}\)
\(=1-\dfrac{1}{64}\)
\(=\dfrac{63}{64}\)
NP
2
N
12 tháng 8 2018
nhân A lên 2A sau đó lấy 2A-A là đc :
2A =1+1/2+.....+1/32
2A-A=(1+1/2+.....+1/32)-(1/2+1/4+.....+1/32+1/64)
A=1-1/64
A=63/64
C
12 tháng 8 2018
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(2A=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\times2\)
\(2A=\frac{1}{2}\times2+\frac{1}{4}\times2+\frac{1}{8}\times2+\frac{1}{16}\times2+\frac{1}{32}\times2+\frac{1}{64}\times2\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\)
\(A=1-\frac{1}{64}\)
\(A=\frac{63}{64}\)
ND
2
1 tháng 3 2017
Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(\Rightarrow2A-A=1-\frac{1}{128}=\frac{127}{128}\)
tính nhanh p/s 1+ 5/4 + 5/8 + 5/16 + 5/32 + 5/64
9 tháng 5 2023
b: A=1/3+1/9+...+1/3^10
=>3A=1+1/3+...+1/3^9
=>A*2=1-1/3^10=(3^10-1)/3^10
=>A=(3^10-1)/(2*3^10)
c: C=3/2+3/8+3/32+3/128+3/512
=>4C=6+3/2+...+3/128
=>3C=6-3/512
=>C=1023/512
d: A=1/2+...+1/256
=>2A=1+1/2+...+1/128
=>A=1-1/256=255/256
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
29 tháng 5 2023
A = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\)+ \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\)+ \(\dfrac{1}{64}\)+ \(\dfrac{1}{128}\)
A\(\times\)2 = 2 + 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\)
A \(\times\) 2 - A = 2 - \(\dfrac{1}{128}\)
A \(\times\)( 2-1) = \(\dfrac{255}{128}\)
A = \(\dfrac{255}{128}\)
BD
29 tháng 5 2023
Gọi \(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\) là T
\(T=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\)
\(2T=2+1+\dfrac{1}{2}+\dfrac{1}{4}+....+\dfrac{1}{64}\)
\(2T-T=\left(2+1+\dfrac{1}{2}+\dfrac{1}{4}+....+\dfrac{1}{64}\right)-\left(1+\dfrac{1}{2}+....+\dfrac{1}{64}+\dfrac{1}{128}\right)\)
\(T=2+\left(1-1\right)+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+....+\left(\dfrac{1}{64}-\dfrac{1}{64}\right)-\dfrac{1}{128}\)
\(T=2+0+0+...-\dfrac{1}{128}\)
\(T=\dfrac{256}{128}-\dfrac{1}{128}\)
\(T=\dfrac{255}{128}\)
AH
Akai Haruma
Giáo viên
31 tháng 3 2023
Lời giải:
$A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}$
$2\times A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}$
$2\times A-A=(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32})-(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64})$
$A=1-\frac{1}{64}=\frac{63}{64}$
Ta có:
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}\)
= \(\frac{1}{1}-\frac{1}{64}=\frac{63}{64}\)
Vậy tổng trên có giá trị là \(\frac{63}{64}\)
cảm ơn