Question

So sánh:1/1×2+1/2×3+...+1/99×100 và 1

Answer 1

Ta có : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{99.100}.\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{100}\)

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

Vậy : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{99.100}< 1\)

Answer 2

Đặt :

\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{99\times100}\)

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

  \(=1-\frac{1}{100}\)

 \(=\frac{99}{100}\)

Vậy  \(A=\frac{99}{100}\)

Vì \(\frac{99}{100}< 1\)nên \(A< 1\)

Học tốt #

Answer 3

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

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

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