B=1/5+1/10+1/20+1/40+............+1/1280.Tìm B
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T
3
1 tháng 8 2018
Tham khảo ở phần Câu hỏi tương tự bạn nhé :
Câu hỏi của Trịnh Thúy An - Toán lớp 5 - Học toán với OnlineMath
1 tháng 8 2018
\(B=\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+...+\frac{1}{1280}\)
\(B=1\cdot\frac{1}{5}+\frac{1}{2}\cdot\frac{1}{5}+\frac{1}{4}\cdot\frac{1}{5}+\frac{1}{8}\cdot\frac{1}{5}+...+\frac{1}{256}\cdot\frac{1}{5}\)
\(B=\frac{1}{5}\cdot\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\right)\)
Đặt \(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\)
\(\Rightarrow2A=2+1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{128}\)
\(\Rightarrow2A-A=2-\frac{1}{256}\)
\(A=2-\frac{1}{256}\)
Thay A vào B
có: \(B=\frac{1}{5}.\left(2-\frac{1}{256}\right)=\frac{1}{5}\cdot\frac{511}{256}=\frac{511}{1280}\)
PK
14 tháng 8 2016
a/ Đặt 1/5= a, ta có:
1/5 + 1/10 + 1/20 + 1/40 + ... + 1/1280
= 1/a + 1/2 x a + 1/4 x a + ... + 1/256 x a
A = 1/a + 1/2 x a + 1/4 x a + ... + 1/256 x a
2 x A = 2/a + 1/a + 1/2 x a + 1/4 x a + ... + 1/128 x a
=> A = 2/a - 1/256 x a = 2/5 - 1/1280 = 511/1280
b/
\(\frac{121}{27}.\frac{54}{11}=\frac{11.11.27.2}{27.11}=11.2=22\)
\(\frac{100}{21}:\frac{25}{126}=\frac{100}{21}.\frac{126}{25}=\frac{25.4.21.6}{21.25}=4.6=24\)
=> \(22< n< 24\)
=> \(n=23\)
13 tháng 8 2016
a) \(A=\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+...+\frac{1}{1280}\)
\(A.2=\frac{2}{5}+\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+...+\frac{1}{640}\)
\(A.2-A=\left(\frac{2}{5}+\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+...+\frac{1}{640}\right)-\left(\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+...+\frac{1}{1280}\right)\)
\(A=\frac{2}{5}-\frac{1}{1280}=\frac{511}{1280}\)
b) \(\frac{121}{27}.\frac{54}{11}< n< \frac{100}{21}:\frac{25}{126}\)
\(22< n< 24\)
=> n = 23
18 tháng 6 2021
A = \(\dfrac{1}{5}+\dfrac{1}{10}+...+\dfrac{1}{1280}\)
= \(\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{20}+...+\dfrac{1}{640}-\dfrac{1}{1280}\)
= \(\dfrac{2}{5}-\dfrac{1}{1280}=\dfrac{511}{1280}\)
19 tháng 6 2021
Giải:
\(\dfrac{1}{5}+\dfrac{1}{10}+\dfrac{1}{20}+\dfrac{1}{40}+...+\dfrac{1}{1280}\)
\(=\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{40}+...+\dfrac{1}{640}-\dfrac{1}{1280}\)
\(=\dfrac{2}{5}-\dfrac{1}{1280}\)
\(=\dfrac{511}{1280}\)
18 tháng 8 2016
\(\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+...+\frac{1}{1280}\)
\(=\frac{1}{5}\left(1+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)\)
\(=\frac{\frac{1}{5}\left(1-\frac{1}{2^9}\right)}{\left(1-\frac{1}{2}\right)}\)
\(=\frac{2}{5}\left(1-\frac{1}{2^9}\right)\)
BH
0
SL
2
1 tháng 7 2018
C = \(\frac{1}{5}\)+\(\frac{1}{10}\)+\(\frac{1}{20}\)+\(\frac{1}{40}\)+\(\frac{1}{80}\)+........+\(\frac{1}{1280}\)
2C = 2 . ( \(\frac{1}{5}\)+\(\frac{1}{10}\)+.......+\(\frac{1}{1280}\))
2C = \(\frac{2}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{10}\)+.....+\(\frac{1}{1280}\)
2C-C = ( \(\frac{2}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{10}\)+......+\(\frac{1}{1280}\)) - (\(\frac{1}{5}\)+\(\frac{1}{10}\)+.....+\(\frac{1}{1280}\))
C . ( 2-1) = \(\frac{2}{5}\)
C = \(\frac{2}{5}\)
Vậy C = \(\frac{2}{5}\)
HN
1 tháng 7 2018
\(C=\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+\frac{1}{80}+........+\frac{1}{1280}\)
\(\Rightarrow2C=2\left(\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+\frac{1}{80}+...........+\frac{1}{1280}\right)\)
\(\Rightarrow2C=\frac{2}{5}+\frac{1}{5}+\frac{1}{10}+.............+\frac{1}{1280}\)
\(\Rightarrow2C-C=\left(\frac{2}{5}+\frac{1}{5}+\frac{1}{10}+............+\frac{1}{1280}\right)-\left(\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+\frac{1}{80}+...........+\frac{1}{1280}\right)\)
\(\Rightarrow C=\frac{2}{5}-\frac{1}{1280}\)
\(\Rightarrow C=\frac{512}{1280}-\frac{1}{1280}\)
\(\Rightarrow C=\frac{511}{1280}\)
Vậy C = \(\frac{511}{1280}\)
TG
30 tháng 10 2016
1/5 + 1/5 - 1/10 + 1/10 - 1/20 + 1/20 - 1/40 + ... + 1/640 - 1/1280
= 1/5 + 1/5 - 1/1280 = 511/1280
A = \(\dfrac{1}{5}\) + \(\dfrac{1}{10}\) + \(\dfrac{1}{20}\) + \(\dfrac{1}{40}\)+ .......+ \(\dfrac{1}{640}\)+ \(\dfrac{1}{1280}\)
2A = \(\dfrac{2}{5}\)+ \(\dfrac{1}{5}\)+ \(\dfrac{1}{10}\) + \(\dfrac{1}{20}\)+ \(\dfrac{1}{40}\)+.........+ \(\dfrac{1}{640}\)
2A - A = \(\dfrac{2}{5}\)- \(\dfrac{1}{1280}\)
A = \(\dfrac{512}{1280}\) - \(\dfrac{1}{1280}\)
A = \(\dfrac{511}{1280}\)