\(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}>\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{100.101}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{101}=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}>\frac{1}{6}\)
\(\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}<\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}<\frac{1}{4}\)
Suy ra: điều cần chứng minh
đặt 1/5^2+1/6^2+,,,+1/100^2=A
*chứng minh A<1/4
ta có: \(\frac{1}{5^2}=\frac{1}{5.5}<\frac{1}{4.5}\)
\(\frac{1}{6^2}=\frac{1}{6.6}<\frac{1}{5.6}\)
...
\(\frac{1}{100^2}=\frac{1}{100.100}<\frac{1}{99.100}\)
\(=>A<\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=>A<\frac{1}{4}-\frac{1}{100}<\frac{1}{4}=>A<\frac{1}{4}\left(1\right)\)
*chứng minh A>1/6
ta có \(\frac{1}{5^2}=\frac{1}{5.5}>\frac{1}{5.6}\)
\(\frac{1}{6^2}=\frac{1}{6.6}>\frac{1}{6.7}\)
...
\(\frac{1}{100^2}=\frac{1}{100.100}>\frac{1}{100.101}\)
\(=>A>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\)
\(=>A>\frac{1}{5}-\frac{1}{101}>\frac{1}{6}=>A>\frac{1}{6}\) (2)
từ (1) và (2)=>1/6<A<1/4 hay 1/6<1/5^2+...+1/100^2<1/4(đpcm)
tick nhé