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\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{899}{30^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{29.31}{30.30}=\frac{1.2.3.....29}{2.3.4.....30}.\frac{3.4.5.....31}{2.3.4.....30}\)
\(=\frac{1}{2}.\frac{31}{30}=\frac{31}{60}\)
\(B=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)......\left(1-\frac{1}{81}\right)\left(1-\frac{1}{100}\right)\)
= \(-\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.......\frac{80}{81}.\frac{99}{100}\)
=\(-\frac{1.3.2.4.3.5..............8.10.9.11}{2^2.3^2.4^2.......10^2}=-\frac{\left(1.2.3.....9\right)\left(3.4.5....11\right)}{2.3.4....10.2.3.4.....10}=-\frac{11}{20}\)
Ta có: \(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)
\(=\left(-\frac{1.3}{2.2}\right).\left(-\frac{2.4}{3.3}\right)...\left(-\frac{99.101}{100.100}\right)\)
\(=-\frac{1}{2}.\frac{101}{100}=-\frac{101}{200}< -\frac{100}{200}=-\frac{1}{2}\)
Vậy \(A< -\frac{1}{2}\)
\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)...\left(\frac{1}{10000}-1\right)\)
\(=\frac{-3}{4}\cdot\frac{-8}{9}\cdot\frac{-15}{16}\cdot...\cdot\frac{-9999}{10000}\)
\(=\frac{-1\cdot3}{2\cdot2}\cdot\frac{-2\cdot4}{3\cdot3}\cdot...\cdot\frac{-99\cdot111}{100.100}\)
\(=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot...\cdot\frac{99\cdot111}{100\cdot100}\)
\(=\frac{\left(1\cdot2\cdot3\cdot4\cdot...\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot6\cdot...\cdot111\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot100\right)^2}\)
\(=\frac{101}{2\cdot100}\)
\(=\frac{101}{200}>\frac{1}{2}\)
Câu A mình làm được nhưng dài quá
B=\(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).............\left(1+\frac{1}{2015}\right)\)
=\(\frac{3}{2}.\frac{4}{3}..............\frac{2016}{2015}\)
=\(\frac{3.4...............2016}{2.3................2015}\)
=\(\frac{2016}{2}=1008\)
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)
\(A=-\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).....\left(1-\frac{1}{100^2}\right)\)
\(A=-\left(\frac{1.3}{2.2}\right).\left(\frac{2.4}{3.3}\right)....\left(\frac{99.101}{100.100}\right)\)
\(A=-\left(\frac{1.2....99}{2.3...100}\right).\left(\frac{3.4....101}{2.3....100}\right)\)
\(A=-\left(\frac{1}{100}\right).\left(\frac{101}{2}\right)\)
\(A=\frac{-101}{200}>\frac{-1}{2}\)
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)
\(=\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}...\frac{100^2-1}{100^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{99.101}{100.100}\)
\(=\frac{\left(1.2.3...99\right)\left(2.3.4...101\right)}{\left(2.3.4...100\right)\left(2.3.4...100\right)}\)
\(=\frac{101}{100}\)
mình lam hơi sai mà cũng khá lac đề, bạn từ bài của mình mà làm bài khác đúng hơn nha hiền
\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{400}-1\right)\)
\(-A=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{400}\right)\)
\(-A=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{399}{400}\)
\(-A=\frac{1\cdot3}{2\cdot2}\cdot\frac{2.4}{3.3}\cdot\frac{3.5}{4.4}\cdot...\cdot\frac{19.21}{20.20}\)
\(-A=\frac{1\cdot2\cdot3\cdot...\cdot19}{2\cdot3\cdot4\cdot...\cdot20}\cdot\frac{3\cdot4\cdot5\cdot...\cdot21}{2\cdot3\cdot4\cdot...\cdot20}\)
\(-A=\frac{1}{20}\cdot\frac{21}{2}=\frac{21}{40}>\frac{20}{40}=\frac{1}{2}\)
\(-A>\frac{1}{2}\Rightarrow A< \frac{1}{2}\)