Sửa đề: \(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\cdot\ldots\cdot\left(1-\frac{1}{1+2+3+\cdots+97}\right)\)
TA có công thức tổng quát: \(1-\frac{1}{1+2+\cdots+n}\)
\(=1-\frac{1}{\frac{n\left(n+1\right)}{2}}\)
\(=1-\frac{2}{n\left(n+1\right)}\)
\(=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)
Ta có: \(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\cdot\ldots\cdot\left(1-\frac{1}{1+2+3+\cdots+97}\right)\)
\(=\frac{\left(2+2\right)\left(2-1\right)}{2\left(2+1\right)}\cdot\frac{\left(3+2\right)\left(3-1\right)}{3\left(3+1\right)}\cdot\ldots\cdot\frac{\left(97+2\right)\left(97-1\right)}{97\left(97+1\right)}\)
\(=\frac{4\cdot1}{2\cdot3}\cdot\frac{5\cdot2}{3\cdot4}\cdot\ldots\cdot\frac{99\cdot96}{97\cdot98}\)
\(=\frac{4\cdot5\cdot\ldots\cdot99}{3\cdot4\cdot\ldots\cdot98}\cdot\frac{1\cdot2\cdot\ldots\cdot96}{2\cdot3\cdot\ldots\cdot97}=\frac{99}{3}\cdot\frac{1}{97}=\frac{33}{97}\)
