Question

[Nguyễn Tuệ Khanh]

I=\(\dfrac{1}{1x3}\)+\(\dfrac{1}{3x5}\)+\(\dfrac{1}{5x7}\)+....\(\dfrac{1}{197x199}\)+\(\dfrac{1}{199x201}\)

Answer 1

[Nguyễn Lê Phước Thịnh]

\(I=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{199\cdot201}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{199\cdot201}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{199}-\dfrac{1}{201}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{200}{201}=\dfrac{100}{201}\)

Answer 2

[Akai Haruma]

Lời giải:

\(2\times I=\frac{2}{1\times 3}+\frac{2}{3\times 5}+\frac{2}{5\times 7}+...+\frac{2}{199\times 201}\)

\(=\frac{3-1}{1\times 3}+\frac{5-3}{3\times 5}+\frac{7-5}{5\times 7}+....+\frac{201-199}{199\times 201}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{199}-\frac{1}{201}\)

\(=1-\frac{1}{201}=\frac{200}{201}\)

\(I=\frac{200}{201}:2=\frac{100}{201}\)