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Cho A = \(\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+......+\frac{3^n-1}{3^n}\) CMR A > n-\(\frac{1}{2}\)
\(A=\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+...+\frac{3^n-1}{3^n}\)
\(=\frac{3-1}{3}+\frac{9-1}{9}+\frac{27-1}{27}+...+\frac{3^n-1}{3^n}\)
\(=\left(\frac{3}{3}-\frac{1}{3}\right)+\left(\frac{9}{9}-\frac{1}{9}\right)+\left(\frac{27}{27}-\frac{1}{27}\right)+.....+\left(\frac{3^n}{3^n}-\frac{1}{3^n}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+....+\frac{1}{3^n}\right)\)
\(=n-\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{3^n}\right)\)
Bây giờ ta chỉ cần chứng minh:\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^n}< \frac{1}{2}\) là xong!
Thật vậy:\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^n}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{n-1}}\)
\(\Rightarrow2B=1-\frac{1}{3^n}\)
\(\Rightarrow B=\frac{1}{2}-\frac{\frac{1}{3^n}}{2}< \frac{1}{2}\)
Ta có:\(A=n-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.....+\frac{1}{3^n}\right)\)
\(>n-\frac{1}{2}\left(đpcm\right)\)(bất đẳng thức đổi chiều)
a)\(VT=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=\frac{1}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right]\)
\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right]\)
\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{3n+2}\right]=\frac{1}{3}\left[\frac{3n+2}{2\left(3n+2\right)}-\frac{2}{2\left(3n+2\right)}\right]\)
\(=\frac{1}{3}\cdot\frac{3n}{6n+4}=\frac{n}{6n+4}=VP\)
b) Ta có: \(\frac{5}{3.7}+\frac{5}{7.11}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)
\(=\frac{5}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\right)\)
\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{4n-1}-\frac{1}{4n+3}\right)\)
\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{4n+3}\right)\)
\(=\frac{5}{4}\left(\frac{4n+3}{12n+9}-\frac{3}{12n+9}\right)\)
\(=\frac{5}{4}.\frac{4n}{12n+9}\)
\(=\frac{5n}{12n+9}\)
( sai đề )
a)\(VT=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=\frac{1}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right]\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\)
\(=\frac{1}{2}-\frac{1}{3n+2}=\frac{3n+2}{2\cdot\left(3n+2\right)}-\frac{2}{2\cdot\left(3n+2\right)}\)
\(=\frac{3n+2-2}{6n+4}=\frac{3n}{6n+4}=VP\)
Với n =1 thì A < 3. Vậy ta phải đi chứng minh A < 3
Giả sử A < 3 đúng với n = k. Ta có:
\(A=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{5}\right)+\left(1+\frac{1}{9}\right)+...+\left(1+\frac{2}{k^2+3k}\right)< 3\)
\(A=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{5}\right)+\left(1+\frac{1}{9}\right)+...+\left(\frac{k^2+3k+2}{k\left(k+3\right)}\right)\)
\(A=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{5}\right)+\left(1+\frac{1}{9}\right)+...+\frac{\left(k+1\right)\left(k+2\right)}{k\left(k+3\right)}\)
Ta phải đi chứng minh A < 3 đúng với n = k +1 tức là phải chứng minh:
\(A=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{5}\right)+\left(1+\frac{1}{9}\right)+...+\frac{\left(k+1\right)\left(k+2\right)}{k\left(k+3\right)}+\left(1+\frac{2}{\left(k+1\right)^2+3\left(k+1\right)}\right)\) \(< 3+\frac{\left(k+2\right)\left(k+3\right)}{\left(k+1\right)\left(k+4\right)}\)
Ta sẽ có:
\(A=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{5}\right)+\left(1+\frac{1}{9}\right)+...+\frac{\left(k+1\right)\left(k+2\right)}{k\left(k+3\right)}+\left(1+\frac{2}{k^2+2k+1+3k+3}\right)\)
\(A=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{5}\right)+\left(1+\frac{1}{9}\right)+...+\frac{\left(k+1\right)\left(k+2\right)}{k\left(k+3\right)}+\frac{k^2+5k+6}{k^2+5k+4}\)
\(A=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{5}\right)+\left(1+\frac{1}{9}\right)+...+\frac{\left(k+1\right)\left(k+2\right)}{k\left(k+3\right)}+\frac{\left(k+2\right)\left(k+3\right)}{\left(k+1\right)\left(k+4\right)}\) \(< 3+\frac{\left(k+2\right)\left(k+3\right)}{\left(k+1\right)\left(k+4\right)}\)
Vậy A đúng với n = k + 1 thì A đúng với n = k
Vậy A < 3 là điều phải chứng minh.
(Phương pháp quy nạp toán học)
