Cho A=\(\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{99}}\)
CMR: A<\(\frac{5}{3}\)
K
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MK
Cho A=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}\)
CMR:0,2<A<0,4
0
9 tháng 7 2017
a, A = \(\frac{1}{2}.\frac{3}{4}.\frac{4}{5}...\frac{99}{100}\)
\(A=\frac{1}{2}.\left(\frac{3.4....99}{4.5...100}\right)\)
\(A=\frac{1}{2}.\left(\frac{3}{100}\right)\)\(\)\(A=\frac{3}{200}\)
\(B=\frac{2}{3}.\frac{4}{5}.\frac{5}{6}...\frac{100}{101}\)
\(B=\frac{2}{3}.\left(\frac{4.5...100}{5.6...101}\right)\)
\(B=\frac{2}{3}.\left(\frac{4}{101}\right)\)
\(B=\frac{8}{303}\)
\(A.B=\frac{8}{303}.\frac{3}{200}\)
\(A.B=\frac{1}{2525}\)
b, A = 1/2 x 3/100
B = 2/3 x 4/101
Ta có : 1 - 2/3 = 1/3; 1 - 1/2 = 1/2
MÀ 1/3 < 1/2 => 2/3 > 1/2 (1)
Ta có : 1 - 3/100 = 97/100
1 - 4/101 = 97/101
Mà 97/101 < 97/100 => 4/101 > 3/100 (2)
Từ (1) và (2) => B > A
9 tháng 7 2017
a,
\(AB=\left[\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}\right]\cdot\left[\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\right]\)
\(AB=\frac{\left[1\cdot3\cdot5\cdot...\cdot99\right]\left[2\cdot4\cdot6\cdot...\cdot100\right]}{\left[2\cdot4\cdot6\cdot8\cdot...\cdot100\right]\left[3\cdot5\cdot7\cdot...\cdot101\right]}=\frac{1\cdot3\cdot5\cdot...\cdot99}{3\cdot5\cdot7\cdot...\cdot101}=\frac{1}{101}\)
b,
1/2 < 2/3
3/4 < 4/5
.............
99/100 < 100/101
=> \(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}< \frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot...\cdot\frac{100}{101}\Leftrightarrow A< B\)
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 < \(\frac{5}{3}\)
\(A=\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{99}}\)
\(A=5\left(\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{99}}\right)\)
\(\frac{A}{5}=\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{99}}\)
\(\frac{4A}{5}=1+\frac{1}{4}+...+\frac{1}{4^{98}}\)
\(\frac{4A}{5}-\frac{A}{5}=\left(1+\frac{1}{4}+...+\frac{1}{4^{98}}\right)-\left(\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{99}}\right)\)
\(\frac{3A}{5}=1-\frac{1}{4^{99}}\Rightarrow A=\frac{5}{3}-\frac{5}{3\cdot4^{99}}< \frac{5}{3}\)