Question

Chứng minh rằng :

S = \(\frac{1}{1x2}\)+\(\frac{1}{2x3}\)+...+\(\frac{1}{99x100}\) < 1

Answer 1

\(S=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

   \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

   \(=1-\frac{1}{100}\)

   \(=\frac{99}{100}< 1\Rightarrowđpcm\)

Answer 2

\(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\)

\(=\frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)

Mà : \(\frac{99}{100}< 1\)

Vậy : S < 1

Answer 3

ủa ? tính nhầm

Answer 4

rồi mình làm đúng hay sai tr ???????