\(s^2\)
A=\(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+....+\frac{1}{4^{1000}}\)và\(\frac{3}{4}\)
K
Khách
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Những câu hỏi liên quan
IM
27 tháng 11 2016
Ta có :
\(C=\frac{1}{4}+\frac{1}{4^2}+.....+\frac{1}{4^{1000}}\)
\(\Rightarrow4C=1+\frac{1}{4}+.....+\frac{1}{4^{1999}}\)
\(\Rightarrow4C-C=\left(1+\frac{1}{4}+.....+\frac{1}{4^{1999}}\right)-\left(\frac{1}{4}+\frac{1}{4^2}+.....+\frac{1}{4^{1000}}\right)\)
\(\Rightarrow3C=1-\frac{1}{4^{1000}}\)
\(\Rightarrow C=\frac{1}{3}-\frac{1}{3.4^{1000}}< \frac{1}{3}\)
=> C < 1 / 3
27 tháng 11 2016
Ta có:
\(C=\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{1000}}\)
\(\Rightarrow4C=1+\frac{1}{4}+...+\frac{1}{4^{999}}\)
\(\Rightarrow4C-C=\left(1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{999}}\right)-\left(\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{999}}+\frac{1}{4^{1000}}\right)\)
\(\Rightarrow3C=1-\frac{1}{4^{1000}}\)
\(\Rightarrow C=\left(1-\frac{1}{4^{1000}}\right).\frac{1}{3}\)
\(\Rightarrow C=\frac{1}{3}-\frac{1}{4^{1000}.3}\)
Mà \(\frac{1}{3}>\frac{1}{3}-\frac{1}{4^{1000}.3}\)
\(\Rightarrow C< \frac{1}{3}\)
Vậy \(C< \frac{1}{3}\)
S
14 tháng 4 2019
\(T=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(T=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)
\(T=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(T=2.\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(T=2.\frac{502}{1005}=\frac{1004}{1005}\)
\(\Rightarrow T=\frac{1004}{1005}\)
S
14 tháng 4 2019
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009+2011}\)
\(A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2009+2011}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)
\(A=\frac{1}{2}.\frac{2010}{2011}\)
\(\Rightarrow A=\frac{1005}{2011}\)
16 tháng 8 2018
\(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\)
\(3A=1+\frac{1}{3}+...+\frac{1}{3^{99}}\)
\(2A=1-\frac{1}{3^{100}}\)
\(A=\frac{1-\frac{1}{3^{100}}}{2}< \frac{1-\frac{1}{3}}{2}=\frac{\frac{2}{3}}{2}=\frac{2}{3}.\frac{1}{2}=\frac{1}{3}< \frac{3}{4}\)
16 tháng 8 2018
vậy \(\frac{1}{^{3^{100}}}\) đâu bạn
mình ko hiểu tại sao lai còn \(\frac{1}{3}\)
NT
15 tháng 6 2017
Tính
a)
\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{9999}{10000}\\ =\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}....\frac{99.101}{100}\\ \)
\(=\left(\frac{1.2.3...99}{2.3...100}\right).\left(\frac{3.4.5...101}{2.3.4...100}\right)\\ =\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)
b)
\(\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{n^2}\\ < \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}\\ \)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{n-1}-\frac{1}{n}\\ =1-\frac{1}{n}< 1\)
PQ
1
HP
11 tháng 7 2016
Đặt \(A=\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+.....+\frac{1}{4^{1000}}\)
\(=>4A=1+\frac{1}{4}+\frac{1}{4^2}+.....+\frac{1}{4^{999}}\)
\(=>4A-A=\left(1+\frac{1}{4}+\frac{1}{4^2}+....+\frac{1}{4^{999}}\right)-\left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+....+\frac{1}{4^{1000}}\right)\)
\(=>3A=1-\frac{1}{4^{1000}}=>A=\frac{1-\frac{1}{4^{1000}}}{3}=\frac{1}{3}-\frac{1}{\frac{4^{1000}}{3}}<\frac{1}{3}\)
Vậy.......................
\(A=\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{1000}}\)
\(\Rightarrow4A=4\left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{1000}}\right)\)
\(\Rightarrow4A=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{998}}+\frac{1}{4^{999}}\)
\(\Rightarrow4A-A=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{998}}+\frac{1}{4^{999}}-\frac{1}{4}-\frac{1}{4^2}-\frac{1}{4^3}-...-\frac{1}{4^{999}}-\frac{1}{4^{1000}}\)
\(\Rightarrow3A=1-\frac{1}{4^{1000}}\)
\(\Rightarrow A=\frac{1-\frac{1}{4^{1000}}}{3}\)
làm tiếp nhé ...okok