1/2+1/4+1/8+...+1/148+1/256
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7 tháng 7 2016
A=1/2+1/4+1/8.....+1/256+1/512
2A=1+1/2+1/4+1/8...1/256
A=(1+1/2+1/4+1/8...1/256)-(1/2+1/4+1/8.....+1/256+1/512)
A=1-1/512
A=511/512
511/512
20 tháng 6 2015
a.\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2015}\right)=\frac{1}{2}.\frac{2}{3}...\frac{2014}{2015}=\frac{1.2.3...2014}{2.3...2015}=\frac{1}{2015}\)
b.\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{128}-\frac{1}{256}=1-\frac{1}{256}=\frac{255}{256}\)
c.\(\frac{5}{2}+\frac{5}{4}+\frac{5}{8}+...+\frac{5}{256}=5\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\right)=5.\frac{255}{256}=\frac{1275}{256}\)
d.14,35+(13,7-13,6).1=14,35+0,1.1=14,35+0,1=14,45
3 tháng 10 2021
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^9}\\ 2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^8}\\ 2A-A=\left(1+\dfrac{1}{2}+...+\dfrac{1}{2^8}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)\\ A=1-\dfrac{1}{2^9}=\dfrac{511}{512}\)
HT
13 tháng 7 2015
Q=\(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{256}\)
Q= \(1+\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+....+\left(\frac{1}{128}-\frac{1}{256}\right)\)
Q=\(1+1+\frac{1}{256}\)
Q=\(\frac{513}{256}\)
NN
1
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
6 tháng 5 2022
Đặt tổng là A
\(2xA=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{64}+\dfrac{1}{128}\)
\(\Rightarrow A=2xA-A=1-\dfrac{1}{256}=\dfrac{255}{256}\)
Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{148}+\frac{1}{256}\)
\(\Rightarrow A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}+\frac{1}{2^8}\)
\(\Rightarrow A=1-\frac{1}{2^7}=1-\frac{1}{128}=\frac{127}{128}\)