Đặt S=1/4+1/16+1/36+...+1/10000

        S= 1/4x(1+1/4+1/9+...+1/2500)

        S= 1/4x(1+1/2x2+1/3x3+...+1/50x50)

S< 1/4x(1+1/1x2+1/2x3+...1/49x50)

S< 1/4x(1+1-1/2+1/2-1/3+....+1/49-1/50)

S< 1/4x(1+1-1/50)

S< 1/4x(2-1/50)<2/4(2/4=1/2)

S< 1/2

S=\(\frac{1}{4}\)(1+\(\frac{1}{2^2}\)+\(\frac{1}{3^2}+...+\frac{1}{50^2}\)

S<\(\frac{1}{4}\)(1+\(\frac{1}{2.1}\)+\(\frac{1}{3.2}+...+\frac{1}{50.49}\))

S<\(\frac{1}{4}\)(1+1−\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\))

S<\(\frac{1}{4}\)(2−\(\frac{1}{50}\))<\(\frac{2}{4}\)=\(\frac{1}{2}\)(đpcm)

Đề phải là \(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+...+\frac{1}{10000}< \frac{1}{2}\) chứ ?

Ta có : \(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+...+\frac{1}{10000}\)

\(=\frac{1}{4}\left(1+\frac{1}{4}+\frac{1}{9}+...+\frac{1}{2500}\right)\)

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

Ta lại có : \(\frac{1}{4}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)< \frac{1}{4}\left(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\right)\)

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

\(=\frac{1}{4}\left(2-\frac{1}{50}\right)=\frac{1}{4}.\frac{99}{50}=\frac{99}{200}\)

Mà \(\frac{99}{200}< \frac{1}{2}\)\(\Rightarrow\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+...+\frac{1}{10000}< \frac{99}{200}< \frac{1}{2}\)

\(\Rightarrow\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+...+\frac{1}{10000}< \frac{1}{2}\) ( đpcm )

\(\text{Đặt BT là A }\Rightarrow A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)

\(\text{Ta có:}\frac{1}{3^2}>\frac{1}{3.4}=\frac{1}{3}-\frac{1}{4}\text{(để lại }\frac{1}{4}\text{ở đầu)}\)

           \(\frac{1}{4^2}>\frac{1}{4.5}=\frac{1}{4}-\frac{1}{5}\)

           .......

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

\(\Rightarrow A>\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{100}-\frac{1}{101}\right)\)

\(\Rightarrow A>\frac{1}{4}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\)

\(\Rightarrow A>\frac{1}{4}+\frac{1}{3}-\frac{1}{101}\Rightarrow A=\frac{7}{12}-\frac{1}{101}=\frac{707-12}{1212}=\frac{695}{1212}>\frac{606}{1212}=\frac{1}{2}\)

\(\Rightarrow A>\frac{1}{2}\)

Mình sai rồi :))

Đặt \(A=\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+...+\dfrac{1}{10000}\)

Ta có:

\(A=\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+...+\dfrac{1}{10000}\)

\(\Rightarrow A=\dfrac{1}{4}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)\)

\(\Rightarrow A< \dfrac{1}{4}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\right)\)

\(\Rightarrow A< \dfrac{1}{4}\left(1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)

\(\Rightarrow A< \dfrac{1}{4}\left(1+1-\dfrac{1}{50}\right)\)

\(\Rightarrow A< \dfrac{1}{4}.\dfrac{99}{50}\)

\(\Rightarrow A< \dfrac{99}{200}< \dfrac{1}{2}\)

Vậy \(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+...+\dfrac{1}{10000}< \dfrac{1}{2}\) (Đpcm)

\(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+...+\dfrac{1}{10000}=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)

\(=\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)< \dfrac{1}{4}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)=\dfrac{1}{4}\left(1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=\dfrac{1}{4}\left(1+1-\dfrac{1}{50}\right)=\dfrac{1}{4}\left(2-\dfrac{1}{50}\right)< \dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+...+\dfrac{1}{10000}< \dfrac{1}{2}\)

\(\frac{1}{2^2}+\frac{1}{4^2}+.....+\frac{1}{100^2}=\frac{1}{2^2}\cdot\left(1+\frac{1}{2^2}+...+\frac{1}{50^2}\right)

Đặt \(M=\frac{1}{4}+\frac{1}{16}+.....+\frac{1}{10000}\)

\(M=\frac{1}{2.2}+\frac{1}{4.4}+\frac{1}{6.6}+......+\frac{1}{100.100}\)

\(=\frac{1}{2.2}\left(1+\frac{1}{2.2}+\frac{1}{3.3}+.....+\frac{1}{50.50}\right)\)

\(< \frac{1}{2.2}\left(1+\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{49.50}\right)\)

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

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

Vậy \(M< \frac{1}{2}\)

99/100

cho mk nhé

Đặt: \(A=\frac{1}{4}+\frac{1}{6}+\frac{1}{36}+\frac{1}{64}+...+\frac{1}{10000}< \frac{1}{2}\)

Ta có: \(A=\frac{1}{4}+\frac{1}{6}+\frac{1}{36}+\frac{1}{64}+...+\frac{1}{10000}\)

\(\Rightarrow A=\frac{1}{4}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)

\(\Rightarrow A< \frac{1}{4}\left(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\right)\)

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

\(\Rightarrow A< \frac{1}{4}\left(1+1-\frac{1}{50}\right)\)

\(\Rightarrow A< \frac{1}{4}.\frac{99}{50}\)

\(\Rightarrow A< \frac{99}{200}< \frac{1}{2}\)

Vậy: \(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+...+\frac{1}{10000}< \frac{1}{2}\left(đpcm\right)\)

Đặt    \(A=\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+...+\frac{1}{10000}\)

\(A=\frac{1}{4}+\frac{1}{4}\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)=\frac{1}{4}+\frac{1}{4}\cdot B\)

Ta có     \(\frac{1}{2^2}< \frac{1}{1\cdot2}=1-\frac{1}{2}\)

\(\frac{1}{3^2}< \frac{1}{2\cdot3}=\frac{1}{2}-\frac{1}{3}\)

\(...\)

\(\frac{1}{50^2}< \frac{1}{49\cdot50}=\frac{1}{49}-\frac{1}{50}\)

\(\Rightarrow B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}< 1\)

\(\Rightarrow A< \frac{1}{4}+\frac{1}{4}\cdot1=\frac{1}{2}\)