Lời giải:
Ta có:
$A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n.(n+1)}=\frac{n}{n+1}$
$=>A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=\frac{99}{99+1}=\frac{99}{100}$
Ta có:$2A=2.\frac{99}{100}=\frac{99}{50}$
Mẫu số chung là:$2.50.101=10100$
$\frac{100}{101}=\frac{10000}{10100}$
$\frac{99}{50}=\frac{19998}{10100}$
$\frac{3}{2}=\frac{15150}{10100}$
Mà:$\frac{10000}{10100}<\frac{19998}{10100}<\frac{15150}{10100}$
$=>\frac{100}{101}<\frac{99}{50}<\frac{3}{2}=>\frac{100}{101}<2A<\frac{3}{2}(đpcm)$
