= 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

\(A=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{99.101}\right)\)

\(\Rightarrow A=\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}.....\frac{99.101+1}{99.101}\)

\(\Rightarrow A=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}.....\frac{10000}{99.101}\)

\(\Rightarrow A=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{100^2}{99.101}\)

\(\Rightarrow A=\frac{\left(2.3.4.....100\right)\left(2.3.4.....100\right)}{\left(1.2.3.....99\right)\left(3.4.5.....101\right)}\)

\(\Rightarrow A=\frac{100.2}{101}=\frac{200}{101}\)

\(A=\left(1+\frac{1}{1\cdot3}\right)\)\(\left(1+\frac{1}{2\cdot4}\right)\)\(\left(1+\frac{1}{3\cdot5}\right)\)\(......\left(1+\frac{1}{99\cdot101}\right)\)

\(=\frac{4}{1\cdot3}\)\(\cdot\frac{9}{2\cdot4}\)\(\cdot\frac{16}{3\cdot5}\)\(\cdot\cdot\cdot\cdot\cdot\frac{10000}{99\cdot101}\)

\(=\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot5}\cdot\cdot\cdot\cdot\frac{100^2}{99\cdot101}\)

\(=\frac{2^2\cdot3^2\cdot4^2\cdot\cdot\cdot100^2}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot\cdot\cdot99\cdot101}\)

\(=\frac{2\cdot3\cdot4\cdot\cdot\cdot\cdot100}{1\cdot2\cdot3\cdot4\cdot\cdot\cdot\cdot99\cdot101}\cdot\frac{2\cdot3\cdot4\cdot\cdot\cdot\cdot100}{3\cdot4\cdot5\cdot\cdot\cdot\cdot99}\)

\(=\frac{1}{101}\cdot200\)

\(=\frac{200}{101}\)

Công thức tống quát:

\(1+\frac{1}{\left(n-1\right)\left(n+1\right)}=1+\frac{1}{n^2-1}=\frac{n^2-1+1}{n^2-1}=\frac{n^2}{n^2-1}\)

Theo đó, ta có:

\(1+\frac{1}{1.3}=1+\frac{1}{\left(2-1\right)\left(2+1\right)}=\frac{2^2}{2^2-1}\)

\(1+\frac{1}{2.4}=1+\frac{1}{\left(3-1\right)\left(3+1\right)}=\frac{3^2}{3^2-1}\)

\(1+\frac{1}{3.5}=\frac{1}{\left(4-1\right)\left(4+1\right)}=\frac{4^2}{4^2-1}\)

\(....................\)

\(1+\frac{1}{2015.2017}=1+\frac{1}{\left(2016-1\right)\left(2016+1\right)}=\frac{2016^2}{2016^2-1}\)

Nhân lần lượt các đẳng thức trên, ta được:

\(S=\frac{\left(2.3.4....2016\right)^2}{\left(2^2-1\right)\left(3^2-1\right)\left(4^2-1\right)...\left(2016^2-1\right)}=\frac{2^2.3^2.4^2...2016^2}{\left(1.3\right)\left(2.4\right)\left(3.5\right)....\left(2015.2017\right)}=\frac{2^2.3^2.4^2...2016^2}{1.2.3^2.4^2.5^2...2014^2.2015^2.2016.2017}=\frac{2.2016}{2017}\)

\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{20.22}\right)\)

\(=\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}.....\frac{20.22+1}{20.22}\)

\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{21^2}{20.22}\)

\(=\frac{\left(2.3.4.....21\right)\left(2.3.4.....21\right)}{\left(1.2.3.....20\right)\left(3.4.5.....22\right)}\)

\(=\frac{21.2}{22}=\frac{42}{22}=\frac{21}{11}\)

Có \(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)..........\)\(\left(1+\frac{1}{2014.2016}\right)\)

=\(\left(\frac{1.3}{1.3}+\frac{1}{1.3}\right)\left(\frac{2.4}{2.4}+\frac{1}{2.4}\right)....\left(\frac{2014.2016}{2014.2016}+\frac{1}{2014.2016}\right)\)

=\(\left(\frac{2^2-1}{1.3}+\frac{1}{2.4}\right)\left(\frac{3^2-1}{2.4}+\frac{1}{2.4}\right)......\left(\frac{2015^2-1}{2014.2016}+\frac{1}{2014.2016}\right)\)

=\(\frac{2.2}{1.3}.\frac{3.3}{2.4}......\frac{2015.2015}{2014.2016}\)

=\(\frac{2.2.3.3.....2015.2015}{1.3.2.4....2014.2015}\)

=\(\frac{\left(2.3...2015\right).\left(2.3.....2015\right)}{\left(1.2....2014\right).\left(3.4.....2016\right)}=\frac{2015.2}{2016}=\frac{4030}{2016}\)

Kết quả là 5/3.Mình ko biết đúng hay sai đâu