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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)
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 < \(\frac{5}{3}\)
Cho C =\(\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{99}}\)
Chứng minh C <\(\frac{5}{3}\)
Bài 2:
M = 1/2.3/4.5/6...99/100
Ta có: \(\frac{a}{b}\) = 1 - \(\frac{b-a}{b}\) (a; b; n ∈ N* và b > a)
\(\frac{a+n}{b+n}\) = 1 - \(\frac{b-a}{b+n}\)
\(\frac{a}{b}\) < \(\frac{a+n}{b+n}\)
Áp dụng công thức trên ta có:
\(\frac12<\frac{1+1}{2+1}=\frac23\)
\(\frac34<\frac{3+1}{4+1}=\frac45\)
\(\frac56\) < \(\frac{5+1}{6+1}\) = \(\frac67\)
............................
\(\frac{99}{100}\) < \(\frac{99+1}{100+1}\) = \(\frac{100}{101}\)
Cộng vế với vế ta có:
M = \(\frac12\).\(\frac34\).\(\frac56\)...\(\frac{99}{100}\) < \(\frac23\).\(\frac45\)..\(\frac{100}{101}\) = N
M < N (đpcm)
b; M.N = \(\frac12\).\(\frac34\).\(\frac56\)...\(\frac{99}{100}\).\(\frac23\).\(\frac45\)..\(\frac{100}{101}\)
M.N = \(\frac{1.3.5\ldots99}{3.5\ldots101}\). \(\frac{2.4.6\ldots100}{2.4.6\ldots100}\)
M.N = 1/100.101
xem link mk
https://olm.vn/hoi-dap/tim-kiem?q=cho+n=1/3-2/3%5E2+3/3%5E3-4/3%5E4+...+99/3%5E99-100/3%5E100+.+Chung+minh+n+%3C+3/16+&id=491985
Ta có:
\(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};\frac{5}{6}< \frac{6}{7};...;\frac{199}{200}< \frac{200}{201}\)
\(\Rightarrow\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{199}{200}< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{200}{201}\)
\(\Rightarrow C< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{200}{201}\)
\(\Rightarrow C^2< \left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{200}{201}\right).\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{199}{200}\right)\)
\(\Rightarrow C^2< \frac{1}{201}\left(dpcm\right)\)
\(C=\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{99}}\)
\(4C=5+\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{98}}\)
\(4C-C=\left(5+\frac{5}{4}+...+\frac{5}{4^{98}}\right)-\left(\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{99}}\right)\)
\(3C=5-\frac{5}{4^{99}}\)
\(C=\frac{5-\frac{5}{4^{99}}}{3}\)
\(C=\frac{5}{3}-\frac{5}{4^{99}.3}< C\)
đpcm
dễ ẹc!!!!!!!!