Question

[Nguyễn Anh Thư]

(1-2/2.3)(1-2/3.4(1-2/4.5)...(1-2/99.100)

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Answer 1

[Pencil]

Ta đặt biểu thức trên là: \(A\)

\(A=\left(1-\frac{2}{2.3}\right)\left(1-\frac{2}{3.4}\right)\left(1-\frac{2}{4.5}\right)\ldots\left(1-\frac{2}{99.100}\right)\)

Ta viết 1 biểu thức chung như sau:

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

\(A=\left(\frac{\left(2+2\right)\left(2-1\right)}{2\left(2+1\right)}\right)\left(\frac{\left(3+2\right)\left(3-1\right)}{3\left(3+1\right)}\right)\left(\frac{\left(4+2\right)\left(4-1\right)}{4\left(4+1\right)}\right)\ldots\left(\frac{\left(99+2\right)\left(99-1\right)}{99\left(99+1\right)}\right)\)

\(A=\frac{\left(4.1\right).\left(5.2\right).\left(6.3\right)\ldots\left(101.98\right)}{\left(2.3\right).\left(3.4\right).\left(4.5\right)\ldots\left(99.100\right)}=\frac{101}{3}\)

Answer 2

[Nguyễn Lê Phước Thịnh]

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

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

\(\left(1-\frac{2}{2\cdot3}\right)\left(1-\frac{2}{3\cdot4}\right)\cdot\ldots\cdot\left(1-\frac{2}{99\cdot100}\right)\)

\(=\frac{\left(2+2\right)\left(2-1\right)}{2\left(2+1\right)}\cdot\frac{\left(3+2\right)\left(3-1\right)}{3\left(3+1\right)}\cdot\ldots\cdot\frac{\left(100+2\right)\left(100-1\right)}{100\left(100+1\right)}\)

\(=\frac{4\cdot1}{2\cdot3}\cdot\frac{5\cdot2}{3\cdot4}\cdot\ldots\cdot\frac{102\cdot99}{100\cdot101}\)

\(=\frac{4\cdot5\cdot\ldots\cdot102}{2\cdot3\cdot\ldots\cdot100}\cdot\frac{1\cdot2\cdot\ldots\cdot99}{3\cdot4\cdot\ldots\cdot101}=\frac{101\cdot102}{2\cdot3}\cdot\frac{1\cdot2}{100\cdot101}\)

\(=\frac{102}{3\cdot100}=\frac{34}{100}=\frac{17}{50}\)