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\(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\\ =\dfrac{1}{5.5}+\dfrac{1}{6.6}+...+\dfrac{1}{100.100}\\ < \dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\\ =\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\)
\(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+....+\dfrac{1}{100^2}>\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{100.101}\\=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{101}\\ =\dfrac{1}{5}-\dfrac{1}{101}\)
1/6<1/5^2+1/6^2+1/7^2+...+1/100^2<1/4
ta có:
(+)1/5^2+1/6^2+1/7^2+...+1/100^2<1/4.5+1/5.6+...+1/99.100
=1/4-1/5+1/5-...+1/99-1/100
=1/4-1/100<1/4
=>1/5^2+1/6^2+1/7^2+...+1/100^2<1/4
(+)1/5^2+1/6^2+1/7^2+...+1/100^2>1/5.6+...+1/99.100
=1/5-1/6+1/6-...+1/99-1/100
=1/5-1/100>1/6
=>1/5^2+1/6^2+1/7^2+...+1/100^2
\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+....+\dfrac{1}{100^2}\\ >\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{100.101}\\ =\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{101}\\ =\dfrac{1}{5}-\dfrac{1}{101}\\ =\dfrac{96}{505}\\ >\dfrac{1}{6}\)
\(\dfrac{1}{5^2}+...+\dfrac{1}{100^2}\\ < \dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+....+\dfrac{1}{99.100}\\ =\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\)
Ta có: \(\dfrac{2}{5^2}+\dfrac{2}{6^2}+\dfrac{2}{7^2}+...+\dfrac{1}{100^2}< \dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{99.100}\)
\(=2.\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{99.100}\right)\)
\(=2.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=2.\left(\dfrac{1}{4}-\dfrac{1}{100}\right)< 2.\dfrac{1}{4}=\dfrac{1}{2}\)
hay \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\)
Vậy ...
Gọi \(A=\dfrac{2}{5^2}+\dfrac{2}{6^2}+\dfrac{2}{7^2}+...+\dfrac{2}{100^2}\)
\(A< \dfrac{2}{4\cdot5}+\dfrac{2}{5\cdot6}+\dfrac{2}{6\cdot7}+...+\dfrac{2}{99\cdot100}\\ A< 2\cdot\left(\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+...+\dfrac{1}{99\cdot100}\right)\\ A< 2\cdot\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ A< 2\cdot\left(\dfrac{1}{4}-\dfrac{1}{100}\right)\\ A< \dfrac{1}{2}-\dfrac{1}{50}\\ A< \dfrac{12}{25}< \dfrac{1}{2}\)
Vậy \(\dfrac{2}{5^2}+\dfrac{2}{6^2}+\dfrac{2}{7^2}+...+\dfrac{2}{100^2}< \dfrac{1}{2}\)
\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{8}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{7}\)
\(\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+\left(-\dfrac{1}{5}+\dfrac{1}{5}\right)+\left(\dfrac{1}{6}-\dfrac{1}{6}\right)+\left(\dfrac{-1}{7}+\dfrac{1}{7}\right)+\dfrac{1}{8}\)
=0+0+0+0+0+0+\(\dfrac{1}{8}\)
=\(\dfrac{1}{8}\)
Đặt: \(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}\)
Ta có: \(\left\{{}\begin{matrix}A>\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{100.101}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{5}-\dfrac{1}{101}>\dfrac{1}{6}\\A< \dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{99.100}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\end{matrix}\right.\)
Vậy \(\dfrac{1}{6}< A< \dfrac{1}{4}\)