\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

Ta có:

\(\frac{1}{51}>\frac{1}{75}\)

\(\frac{1}{52}>\frac{1}{75}\)

......................

\(\frac{1}{75}=\frac{1}{75}\)

\(\Rightarrow\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}>\frac{1}{75}+\frac{1}{75}+...+\frac{1}{75}=25.\frac{1}{75}=\frac{1}{3}\)(1)

Ta có:

\(\frac{1}{76}>\frac{1}{100}\)

\(\frac{1}{77}>\frac{1}{100}\)

........................

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

\(\Rightarrow\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=25.\frac{1}{100}=\frac{1}{4}\)(2)

Từ (1) và (2) ta có:

\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}+\frac{1}{76}+...+\frac{1}{100}>\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)

\(\Rightarrow\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}>\frac{7}{12}\)(5)

Ta có:

\(\frac{1}{51}=\frac{1}{51}\)

\(\frac{1}{52}< \frac{1}{51}\)

...................

\(\frac{1}{75}< \frac{1}{51}\)

\(\Rightarrow\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}< \frac{1}{51}+\frac{1}{51}+...+\frac{1}{51}=25.\frac{1}{51}< 25.\frac{1}{50}=\frac{1}{2}\)(3)

Ta có:

\(\frac{1}{76}=\frac{1}{76}\)

\(\frac{1}{77}< \frac{1}{76}\)

...................

\(\frac{1}{100}< \frac{1}{76}\)

\(\Rightarrow\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}< \frac{1}{76}+\frac{1}{76}+...+\frac{1}{76}=25.\frac{1}{76}< 25.\frac{1}{75}=\frac{1}{3}\)(4)

Từ (3) và (4) ta có:

\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}+\frac{1}{76}+...+\frac{1}{100}>\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)

\(\Rightarrow\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}< \frac{5}{6}\)(6)

Từ (5) và (6) 

\(\Rightarrow\frac{7}{12}< \frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}< \frac{5}{6}\)

                                                            đpcm

Tham khảo nhé~

Ta có

\(A=\frac{2-1}{1.2}+\frac{4-3}{3.4}+\frac{6-5}{5.6}+...+\frac{100-99}{99.100}\)

\(A=\frac{2}{1.2}-\frac{1}{1.2}+\frac{4}{3.4}-\frac{3}{3.4}+\frac{6}{5.6}-\frac{5}{5.6}+...+\frac{100}{99.100}-\frac{99}{100.99}\)

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

<=>A=1-1/100=99/100

=>7/12<A<5/6(bấm máy tính là biết)

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{99}+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

\(A=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\)

\(A=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}\right)+\left(\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}\right)\)

Ta có : \(\frac{1}{51}>\frac{1}{52}>...>\frac{1}{75}\), \(\frac{1}{76}>\frac{1}{77}>...>\frac{1}{100}\)nên :

\(A>\frac{1}{75}.25+\frac{1}{100}.25=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)

\(A< \frac{1}{51}.25+\frac{1}{76}.25< \frac{1}{50}.25+\frac{1}{75}.25=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)

Vậy \(\frac{7}{12}< A< \frac{5}{6}\)

+A=1/(1.2)+1/(3.4)+...+1/(99.100) 
=1/1-1/2+1/3-1/4+....+1/99-1/100 
=1/2+1/3-1/4+1/5-1/6+1/7+...-1/98+1/99... 
=(1/2+1/3)+(1/5-1/4)+(1/7-1/6)+..+(1/9... 
=5/6-(1/4.5+1/6.7+..1/98.99+1/100)<5/6 
do -(1/4.5+1/6.7+..1/98.99+1/100)<0 
+A=1/(1.2)+1/(3.4)+...+1/(99.100) 
=1/2+1/12+1/(5.6)+...+1/(99.100) 
=7/12+[1/(5.6)+...1/(99.100)] 
>7/12 do [1/(5.6)+...1/(99.100)]>0

CM cai j cay

//Chứng minh gì thế=)) < 7/12 hay > 7/12 hay = 7/12

\(C=\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+....+\frac{99.100-1}{100!}\)

\(\Rightarrow C=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)

\(\Rightarrow C=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)

\(\Rightarrow C=\left(2+\frac{3.4}{4!}+\frac{4.5}{5!}+....+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{10!}\right)\)

\(\Rightarrow C=\left(2+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)

\(\Rightarrow C=2-\frac{1}{99!}-\frac{1}{100!}< 2\Rightarrow C< 2\)

\(b,C=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+....+\frac{19}{9^2.10^2}\)

\(\Rightarrow C=\frac{3}{\left(1.2\right)\left(1.2\right)}+\frac{5}{\left(2.3\right)\left(2.3\right)}+...+\frac{19}{\left(9.10\right)\left(9.10\right)}\)

\(\Rightarrow C=\frac{3}{1.2}.\frac{1}{1.2}+\frac{5}{2.3}.\frac{1}{2.3}+....+\frac{19}{9.10}.\frac{1}{9.10}\)

\(\Rightarrow C=\left(1+\frac{1}{2}\right)\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}+\frac{1}{3}\right)\left(\frac{1}{2}-\frac{1}{3}\right)+....+\left(\frac{1}{9}+\frac{1}{10}\right)\left(\frac{1}{9}-\frac{1}{10}\right)\)

\(\Rightarrow C=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+....+\frac{1}{81}-\frac{1}{90}\)

\(\Rightarrow C=1-\frac{1}{90}< 1\Rightarrow C< 1\)

ta có: A=\(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}=>A=\left(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+..+\frac{1}{75}\right)+\left(\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}\right)\)

\(\frac{1}{51}>\frac{1}{52}>\frac{1}{53}>...>\frac{1}{75};\frac{1}{76}>\frac{1}{77}>...>\frac{1}{100}\)

do đó:\(A>\frac{1}{75}.25+\frac{1}{100}.25=>A>\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)  (1)

lại có: \(A<\frac{1}{51}.25+\frac{1}{76}.25<\frac{1}{50}.25+\frac{1}{75}.25=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)   (2)

từ (1) và (2)=>7/12<A<5/6(đpcm)

Ta có vế trái:

\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

                                                           \(=\left(1+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)

                                                           \(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)

                                                           \(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)

                                                           \(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}=\)vế phải

Vậy\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\RightarrowĐpcm\)