Question

[6a01dd_nguyenphuonghoa.]

(1+1/2).(1+1/3).(1+1/4)...(1+1/99)

GIẢI CHI TIẾT GIÚP MÌNH VỚI Ạ. CẢM ƠN MỌI NGƯỜI!

Answer 1

[HT.Phong (9A5)]

\(\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{99}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot...\cdot\dfrac{97}{98}\cdot\dfrac{98}{99}\)

\(=\dfrac{1\cdot2\cdot3\cdot...\cdot98}{2\cdot3\cdot4\cdot...\cdot99}\)

\(=\dfrac{1}{99}\)

Answer 2

[Nguyễn Ngọc Anh Minh]

\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}.....\dfrac{99}{98}.\dfrac{100}{99}=\dfrac{100}{2}=50\)