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A=1/1.2+1/12+...+1/99.100
A=7/12+...1/99.100
Suy ra A>7/12 (1)
A=1-1/2+1/3-1/4+...+1/99-1/100
A=(1/2+1/3)-(1/4-...+1/100)
A=5/6-(1/4-...+1/100)
suy ra A<5/6 (2)
Vậy 7/12<A<5/6
chắc chắn đúng
Lê Tùng lâm bài của bạn chưa đúng vì
A = \(\frac{1}{1.2}+\frac{1}{3.4}+.....+\frac{1}{99.100}\)
Chứ không phải là: \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{98.99}+\frac{1}{99.100}\)
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2019.2020}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(=1-\frac{1}{2020}< 1\)
Vậy \(A< 1\left(đpcm\right)\)
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(\Leftrightarrow B< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Leftrightarrow B< \frac{1}{4}+\frac{1}{2}-\frac{1}{50}\)
\(\Leftrightarrow B< \frac{1}{4}+\frac{1}{2}\)
\(\Leftrightarrow B< \frac{3}{4}\left(đpcm\right)\)
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)
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{99}+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(A=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\)
\(A=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}\right)+\left(\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}\right)\)
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