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= 4/1.3 x 9/2.4 x 16/3.5 x...x 10000/99.101
= 2.2/1.3 x 3.3/2.4 x 4.4/3.5 x..x 100.100/99.101
= (2.3.4. ... 100/1.2.3. .... 99) x (2.3.4. ... .100/3.4.5. ... .101)
= 100.2/101
=200/101
\(A=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{99.101}\right)\)
\(\Rightarrow A=\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}.....\frac{99.101+1}{99.101}\)
\(\Rightarrow A=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}.....\frac{10000}{99.101}\)
\(\Rightarrow A=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{100^2}{99.101}\)
\(\Rightarrow A=\frac{\left(2.3.4.....100\right)\left(2.3.4.....100\right)}{\left(1.2.3.....99\right)\left(3.4.5.....101\right)}\)
\(\Rightarrow A=\frac{100.2}{101}=\frac{200}{101}\)
Có \(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)..........\)\(\left(1+\frac{1}{2014.2016}\right)\)
=\(\left(\frac{1.3}{1.3}+\frac{1}{1.3}\right)\left(\frac{2.4}{2.4}+\frac{1}{2.4}\right)....\left(\frac{2014.2016}{2014.2016}+\frac{1}{2014.2016}\right)\)
=\(\left(\frac{2^2-1}{1.3}+\frac{1}{2.4}\right)\left(\frac{3^2-1}{2.4}+\frac{1}{2.4}\right)......\left(\frac{2015^2-1}{2014.2016}+\frac{1}{2014.2016}\right)\)
=\(\frac{2.2}{1.3}.\frac{3.3}{2.4}......\frac{2015.2015}{2014.2016}\)
=\(\frac{2.2.3.3.....2015.2015}{1.3.2.4....2014.2015}\)
=\(\frac{\left(2.3...2015\right).\left(2.3.....2015\right)}{\left(1.2....2014\right).\left(3.4.....2016\right)}=\frac{2015.2}{2016}=\frac{4030}{2016}\)
\(=\frac{4}{3}.\frac{9}{8}...\frac{4060225}{4060224}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}...\frac{2015.2015}{2014.2016}\)
\(=\frac{2.2.3.3...2015.2015}{1.3.2.4...2014.2016}\)
\(=\frac{2.3...2015}{1.2...2014}.\frac{2.3...2015}{3.4...2016}\)
\(=2015.\frac{2}{2016}\)
\(=2015.\frac{1}{1008}\)
\(=\frac{2015}{1008}\)
Gọi n là số tự nhiên khác 0, ta có:
\(1+\frac{1}{n\left(n+2\right)}=\frac{n\left(n+2\right)}{n\left(n+2\right)}+\frac{1}{n\left(n+2\right)}\)
\(=\frac{n^2+2n+1}{n\left(n+2\right)}=\frac{n^2+n+n+1}{n\left(n+2\right)}\)
\(=\frac{n\left(n+1\right)+\left(n+1\right)}{n\left(n+2\right)}=\frac{\left(n+1\right)\left(n+1\right)}{n\left(n+2\right)}\)
\(=\frac{n+1}{n}.\frac{n+1}{n+2}\)
Vậy \(1+\frac{1}{n\left(n+2\right)}=\frac{n+1}{n}.\frac{n+1}{n+2}\)
Áp dụng vào biểu thức A ta có:
A = \(\left(\frac{1+1}{1}.\frac{1+1}{1+2}\right).\left(\frac{2+1}{2}.\frac{2+1}{2+2}\right)\)\(.\left(\frac{3+1}{3}.\frac{3+1}{3+2}\right)....\left(\frac{2017+1}{2017}.\frac{2017+1}{2017+2}\right)\)
A = \(\frac{2}{1}.\frac{2}{3}.\frac{3}{2}.\frac{3}{4}.\frac{4}{3}.\frac{4}{5}.\frac{5}{4}...\frac{2018}{2017}.\frac{2018}{2019}\)
A = \(\frac{2}{1}.1.1.1...1.\frac{2018}{2019}\)
A = \(2.\frac{2018}{2019}\)
Vì \(\frac{2018}{2019}< 1\)(vì 2018 < 2019)
Nên \(A=2.\frac{2018}{2019}< 2.1=2\)
Anh em trả lời giúp mình với !