Ta có :

\(\dfrac{1}{5^2}< \dfrac{1}{4.5}\)

\(\dfrac{1}{6^2}< \dfrac{1}{5.6}\)

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

\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)

\(\Leftrightarrow\dfrac{1}{5^2}+\dfrac{1}{6^2}+........+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+.....+\dfrac{1}{99.100}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\left(1\right)\)

Lại có :

\(\dfrac{1}{5^2}>\dfrac{1}{5.6}\)

\(\dfrac{1}{6^2}>\dfrac{1}{6.7}\)

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

\(\dfrac{1}{100^2}>\dfrac{1}{100.101}\)

\(\Leftrightarrow\dfrac{1}{5^2}+\dfrac{1}{6^2}+......+\dfrac{1}{100^2}>\dfrac{1}{5.6}+\dfrac{1}{6.7}+.....+\dfrac{1}{100.101}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+.....+\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{5}-\dfrac{1}{101}>\dfrac{1}{6}\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)

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 \(\dfrac{1}{6}< A< \dfrac{1}{4}\)

a)\(A=\dfrac{1}{2^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2^2-1}+\dfrac{1}{4^2-1}+...+\dfrac{1}{100^2-1}\)

\(A< \dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{99\cdot101}\)

\(A< \dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(A< \dfrac{1}{2}\cdot\left(1-\dfrac{1}{101}\right)=\dfrac{1}{2}\cdot\dfrac{100}{101}=\dfrac{50}{101}< \dfrac{50}{100}=\dfrac{1}{2}\)

Vậy \(A< \dfrac{1}{2}\)

b)B=\(\dfrac{3}{4}+\dfrac{8}{9}+...+\dfrac{2499}{2500}\)

49-B=\(\dfrac{1}{4}+\dfrac{1}{9}+...+\dfrac{1}{2500}=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\)

\(49-B< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(49-B< 1-\dfrac{1}{50}< 1\Leftrightarrow49< 1+B\Leftrightarrow B>48\)(ĐPCM)

b) Đặt :

\(A=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+............+\dfrac{2499}{2500}\)

\(\Rightarrow A=\dfrac{4}{4}-\dfrac{1}{4}+\dfrac{9}{9}-\dfrac{1}{9}+.........+\dfrac{2500}{2500}-\dfrac{1}{2500}\)

\(A=1-\dfrac{1}{2^2}+1-\dfrac{1}{3^2}+...........+1-\dfrac{1}{50^2}\)

\(A=\left(1+1+....+1\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+......+\dfrac{1}{50^2}\right)\)(\(49\) chữ số \(1\))

\(A=49-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+........+\dfrac{1}{50^2}\right)\)

Lại có :

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+.....+\dfrac{1}{50^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{49.50}\)

Mà :

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{49.50}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=1-\dfrac{1}{50}< 1\)

\(\Rightarrow-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+......+\dfrac{1}{50^2}\right)>-1\)

\(\Rightarrow49-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+............+\dfrac{1}{50^2}\right)>49-1\)\(=48\)

\(\Rightarrow A>48\) \(\rightarrowđpcm\)

b\()\)

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2.3 + 1/3.4 +... + 1/99.100

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/99 + 1/100

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/100

1/2^2 + 1/3^2 +... + 1/100^2 < 3/4 - 1/100 < 3/4

Tương tự như vậy với câu a\()\)

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2.3 + 1/3.4 +... + 1/99.100

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/99 + 1/100

1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/100

1/2^2 + 1/3^2 +... + 1/100^2 < 3/4 - 1/100 < 1/2

Ta có: \(K=\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}< \dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)

\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{3}-\dfrac{1}{100}< \dfrac{1}{3}\) (1)

\(K=\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}>\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{100.101}\)

\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{4}-\dfrac{1}{101}>\dfrac{1}{5}\) (2)

Từ (1), (2) \(\Rightarrow\dfrac{1}{5}< K< \dfrac{1}{3}\left(đpcm\right)\)

Bạn nào có thể giải giùm mình bài toán sau:

so sánh : 11¹³ và 13¹¹

\(\dfrac{1}{1\cdot2}>\dfrac{1}{2^2}>\dfrac{1}{2\cdot3},\dfrac{1}{2\cdot3}>\dfrac{1}{3^2}>\dfrac{1}{3\cdot4},...,\dfrac{1}{8\cdot9}>\dfrac{1}{9^2}>\dfrac{1}{9\cdot10}\)

\(\Rightarrow\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{8\cdot9}>\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{9^2}>\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\) \(\Rightarrow1-\dfrac{1}{9}>A>\dfrac{1}{2}-\dfrac{1}{10}\) \(\Rightarrow\dfrac{8}{9}>A>\dfrac{2}{5}\)

\(a,P=\dfrac{1}{\left(2+1\right)\left(2+1-1\right):2}+\dfrac{1}{\left(3+1\right)\left(3+1-1\right):2}+...+\dfrac{1}{\left(2017+1\right)\left(2017+1-1\right):2}\\ P=\dfrac{1}{2\cdot3:2}+\dfrac{1}{3\cdot4:2}+...+\dfrac{1}{2017\cdot2018:2}\\ P=2\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2017\cdot2018}\right)\\ P=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}\right)\\ P=2\left(\dfrac{1}{2}-\dfrac{1}{2018}\right)=2\cdot\dfrac{504}{1009}=\dfrac{1008}{1009}\)

\(b,\) Ta có \(\dfrac{1}{4^2}< \dfrac{1}{2\cdot4};\dfrac{1}{6^2}< \dfrac{1}{4\cdot6};...;\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{\left(2n-2\right)2n}\)

\(\Leftrightarrow VT< \dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{\left(2n-2\right)2n}\\ \Leftrightarrow VT< \dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{\left(2n-2\right)2n}\right)\\ \Leftrightarrow VT< \dfrac{1}{2}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2n-2}-\dfrac{1}{2n}\right)\\ \Leftrightarrow VT< \dfrac{1}{2}\left(1-\dfrac{1}{2n}\right)< \dfrac{1}{2}\cdot\dfrac{1}{2}=\dfrac{1}{4}\)