\(M=\dfrac{1}{2^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)

Ta thấy \(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2};\dfrac{1}{4^2}< \dfrac{1}{3\cdot4};...;\dfrac{1}{100^2}< \dfrac{1}{99\cdot100}\)

\(\Rightarrow M< \dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\\ =1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\\ =\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\\ =\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)-1-\dfrac{1}{2}-...-\dfrac{1}{50}\\ =\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}< \dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}\left(50.số\right)=\dfrac{50}{50}=1\)

Vậy \(M< 1\)

Mình chỉ so sánh với 1 được thôi à :((

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

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

\(...\)

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

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

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

\(Đ\text{ặt }S=\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+....+\frac{1}{10000}\)

\(S=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)

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

Ta có :

\(\frac{1}{2^2}< \frac{1}{1\cdot2};\text{ }\frac{1}{3^2}< \frac{1}{2\cdot3};\text{ }...;\text{ }\frac{1}{50^2}< \frac{1}{49\cdot50}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}\)

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

\(=1-\frac{1}{50}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1\Rightarrow1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+1=2\)

\(\Rightarrow\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)< \frac{1}{2^2}\cdot2\)

\(\Rightarrow S< \frac{1}{2}\) (ĐPCM)

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

\(\Rightarrow A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{100^2}\)

\(\Rightarrow4A=1+\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{50^2}\)

\(\Rightarrow4A< 1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}\)

\(\Rightarrow4A=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(\Rightarrow4A< 2-\frac{1}{50}< 2\)

\(\Rightarrow4A< 2\Rightarrow A< \frac{2}{4}=\frac{1}{2}\)

=>a<1/2

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)

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

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

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

ra ba\o nhyieu