Cho \(C=\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{99}}.\)\(CMR:C<\frac{5}{3}\)
Triều on ko?
K
Khách
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26 tháng 12 2017
\(C=\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{99}}\)
\(C=5.\left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{99}}\right)\)
đặt \(A=\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{99}}\)
\(4A=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{98}}\)
\(4A-A=\left(1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{98}}\right)-\left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{99}}\right)\)
\(3A=1-\frac{1}{4^{99}}\)< 1
\(\Rightarrow A=\frac{1-\frac{1}{4^{99}}}{3}< \frac{1}{3}\)
suy ra \(C=5.\left(\frac{1-\frac{1}{4^{99}}}{3}\right)< 5.\frac{1}{3}=\frac{5}{3}\)
TN
26 tháng 12 2017
Ta có :
\(C=\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{99}}\)
\(\Rightarrow4.C=4\left(\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{99}}\right)\)
\(\Rightarrow4C=5+\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{98}}\)
\(\Rightarrow4C-C=5+\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{98}}-\frac{5}{4}-\frac{5}{4^2}-...-\frac{5}{4^{99}}\)
\(\Rightarrow3C=\)\(5-\frac{5}{4^{99}}=5\left(1-\frac{1}{4^{99}}\right)\)
\(\Rightarrow C=\frac{5}{3}.\left(1-\frac{1}{4^{99}}\right)< \frac{5}{3}\left(đpcm\right)\)
Kb vs k cho mình nhé!
1 tháng 5 2016
4C=\(5+\frac{5}{4}+\frac{5}{4^2}+.......+\frac{5}{4^{98}}\)
4C-C=\(5-\frac{5}{4^{99}}\)
3C=\(5-\frac{5}{4^{99}}<5\)
\(\Rightarrow C<\frac{5}{3}\)
S
24 tháng 6 2017
C = \(\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{99}}\)
= \(5\left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{99}}\right)\)
Đặt A = \(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{99}}\)
4A = \(1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{99}}\)
4A - A = \(\left(1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{99}}\right)-\left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{99}}\right)\)
3A = \(1-\frac{1}{4^{99}}< 1\)
=> A < \(\frac{1}{3}\) (1)
Thay (1) vào C ta được:
\(C< 5\cdot\frac{1}{3}=\frac{5}{3}\)(đpcm)
OD
24 tháng 6 2017
Ta có:\(\frac{5}{4}\)< \(\frac{5}{3}\)Mà C = \(\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{99}}\)<\(\frac{5}{4}\)
\(\Rightarrow\)C < \(\frac{5}{3}\)
TX
Cho C =\(\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{99}}\)
Chứng minh C <\(\frac{5}{3}\)
0
29 tháng 1 2017
\(A=\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{99}}\)
\(A=5\left(\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{99}}\right)\)
\(\frac{A}{5}=\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{99}}\)
\(\frac{4A}{5}=1+\frac{1}{4}+...+\frac{1}{4^{98}}\)
\(\frac{4A}{5}-\frac{A}{5}=\left(1+\frac{1}{4}+...+\frac{1}{4^{98}}\right)-\left(\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{99}}\right)\)
\(\frac{3A}{5}=1-\frac{1}{4^{99}}\Rightarrow A=\frac{5}{3}-\frac{5}{3\cdot4^{99}}< \frac{5}{3}\)
13 tháng 8 2018
\(C=\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{99}}\)
\(4C=5+\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{98}}\)
\(4C-C=\left(5+\frac{5}{4}+...+\frac{5}{4^{98}}\right)-\left(\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{99}}\right)\)
\(3C=5-\frac{5}{4^{99}}\)
\(C=\frac{5-\frac{5}{4^{99}}}{3}\)
\(C=\frac{5}{3}-\frac{5}{4^{99}.3}< C\)
đpcm
Nguyễn Huy Hải, papa Dung và mama Triều hả?=.=