Question

[vũ huy nhật]

Tính giúp mình bài toán này nhé

Tính H 

H=(1+1/1.3).(1+1/2.4).(1+1/3.5).....(1+1/99.100)

Answer 1

[Nguyễn Lê Phước Thịnh]

Sửa đề: \(H=\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\cdot...\cdot\left(1+\dfrac{1}{99\cdot101}\right)\)

\(=\left(1+\dfrac{1}{2^2-1}\right)\left(1+\dfrac{1}{3^2-1}\right)\cdot...\cdot\left(1+\dfrac{1}{100^2-1}\right)\)

\(=\dfrac{2^2}{2^2-1}\cdot\dfrac{3^2}{3^2-1}\cdot...\cdot\dfrac{100^2}{100^2-1}\)

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

\(=\dfrac{100}{1}\cdot\dfrac{2}{101}=\dfrac{200}{101}\)

Answer 2

[Hoàng Trung Hải]

khó quá