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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(1+\frac{1}{3}\right)\left(1+\frac{1}{8}\right)\left(1+\frac{1}{15}\right)...\left(1+\frac{1}{120}\right)\)
\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}...\frac{121}{120}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{11.11}{10.12}\)
\(=\frac{2.2.3.3.4.4...11.11}{1.3.2.4.3.5...10.12}\)
\(=\frac{\left(2.3.4...11\right)\left(2.3.4...11\right)}{\left(1.2.3...10\right).\left(3.2.5...12\right)}\)
\(=\frac{11.2}{1.12}\)
\(=\frac{11}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(1+1/3)(1+1/8)(1+1/15)...(1+1/9603)=4/3 . 9/8 . 16/15 ... 9604/9603
= (2.2)/(1.3) . (3.3)/(2.4) . (4.4)/(3.5) ... (98.98)/(97.99)
=(2.2.3.3.4.4...98.98)/(1.3.2.4.3.5...97.99)
=(2.3.4...98)/(1.2.3...97) . (2.3.4..98)/(3.4.5...99)
=98/1 .2/99 =169/99 .
![](https://rs.olm.vn/images/avt/0.png?1311)
Nhận xét :\(1-\frac{1}{n^2}=\frac{n^2-1}{n^2}=\frac{\left(n+1\right)\left(n-1\right)}{n^2}\)
Do đó : \(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right).....\left(1-\frac{1}{100^2}\right)\)
\(=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}.........\frac{99.101}{100^2}\)
\(=\frac{\left(1.2.3....99\right)\left(3.4.5....101\right)}{\left(2.3.4.....100\right)\left(2.3.4.....100\right)}\)
\(=\frac{101}{100.2}\)
\(=\frac{101}{200}\)
\(\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{8}\right)\left(1+\dfrac{1}{15}\right)...\left(1+\dfrac{1}{120}\right)\)
\(=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}...\dfrac{121}{120}=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}.\dfrac{4.4}{3.5}.\dfrac{11.11}{10.12}\)
\(=\dfrac{2}{1}.\dfrac{11}{12}=\dfrac{11}{6}\)