Tính
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
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27 tháng 8 2018
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=\frac{64}{128}+\frac{32}{128}+\frac{16}{128}+\frac{8}{128}+\frac{4}{128}+\frac{2}{128}\)
\(=\frac{126}{128}=\frac{63}{64}\)
8 tháng 7 2017
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\)
\(=\frac{1+1+1+1+1+1+1}{2}\)
\(=\frac{7}{2}\)
DP
8 tháng 7 2017
Đặt \(T=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(T=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+...+\left(\frac{1}{64}-\frac{1}{128}\right)\)
\(\Rightarrow T=1-\frac{1}{128}=\frac{127}{128}\)
5 tháng 6 2018
Theo đề bài ta có :
\(2B=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{128}\)
\(\Leftrightarrow2B-B=\left(1+\frac{1}{2}+...+\frac{1}{128}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\right)\)
\(\Leftrightarrow B=1-\frac{1}{256}\)
\(\Leftrightarrow B=\frac{255}{256}\)
5 tháng 6 2018
\(B=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+..+\frac{1}{256}\)
\(\Rightarrow B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+..+\frac{1}{2^8}\)
\(\Rightarrow2B=1+\frac{1}{2}+\frac{1}{2^2}+..+\frac{1}{2^7}\)
\(\Rightarrow2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\right)\)
\(\Rightarrow B=1-\frac{1}{2^8}\)
HT
8 tháng 8 2015
Đặt A = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
2A = \(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
2A - A = \(1-\frac{1}{64}\)
=> A = \(\frac{63}{64}\)
HT
1 tháng 8 2015
Cách 1:
Đặt A = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\)
2A = \(1+\frac{1}{2}+\frac{1}{4}+....+\frac{1}{64}\)
A = 2A - A = \(1-\frac{1}{128}\)
=> A = \(\frac{127}{128}\)
Cách 2:
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\)
= \(\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+...+\left(\frac{1}{64}-\frac{1}{128}\right)\)
= \(1-\frac{1}{128}\)
= \(\frac{127}{128}\)
VQ
6 tháng 5 2017
1/2 - 1/4 + 1/4 - 1/8 + 1/8 - 1/16 + 1/16 - 1/32 + 1/32 - 1/64 + 1/64 - 1/128
Gạch 1/4 với 1/4 , 1/8 với 1/8 , 1/16 với 1/16 , 1/32 với 1/32 , 1/64 với 1/64
Còn 1/2 - 1/128 = 63/128
Đúng thì k cho mình
L
1 tháng 5 2019
\(A\cdot2=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...+\frac{1}{256}\right)\cdot2\)
\(=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...+\frac{1}{128}\)
\(A\cdot2-A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...+\frac{1}{128}-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\right)\)
\(A=1-\frac{1}{256}=\frac{255}{256}\)
1 tháng 5 2019
\(A=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^7}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^7}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)\)
\(A=1-\frac{1}{2^8}\)
\(A=\frac{2^8-1}{2^8}\)
\(A=\frac{255}{256}\)
SA
4 tháng 7 2018
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}=\frac{32}{64}+\frac{16}{64}+\frac{8}{64}+\frac{4}{64}+\frac{2}{32}+\frac{1}{64}\)
\(\frac{32+16+8+4+2}{64}=\frac{62}{64}=\frac{31}{32}\)
Tk mh nhé , mơn nhìu !!!
~ HOK TỐT ~
4 tháng 7 2018
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)\(+\frac{1}{64}\)
= 32/64 + 16/64 + 8/64 + 4/64 + 2/64 + 1/64
= 63/64
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}\)
\(=1-\frac{1}{64}\)
\(=\frac{63}{64}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{32}-\frac{1}{64}\)
\(=1-\frac{1}{64}\)
\(=\frac{64}{64}-\frac{1}{64}=\frac{63}{64}\)