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

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

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

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

Ta có: \(VT=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)

\(4VT=\dfrac{1}{2^2:2^2}+\dfrac{1}{4^2:2^2}+\dfrac{1}{6^2:2^2}+...+\dfrac{1}{100^2:2^2}\)

\(4VT=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\)

Lại có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

\(...\)

\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)

\(\Rightarrow4VT-1< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)(*)

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

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

Từ (*) và (**) \(\Rightarrow4VT< 2-\dfrac{1}{50}\)

\(\Rightarrow VT< \dfrac{1}{2}-\dfrac{1}{200}< VP\Rightarrow\) đpcm

b) Ta có: \(2VT=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\)

\(2VT+VT=\left(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\right)\)

\(3VT=1-\dfrac{1}{64}< 1\)

\(\Rightarrow VT< \dfrac{1}{3}\) (đpcm)

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

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

\(...\)

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

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

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

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

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 :((

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 :))

ko bít làm thì thôi đi dễ quá mà

số số hạng của dãy số:

(10000-1):3+1=3334

tổng của dãy số là:

(10000+1).3334:2=16671667

k nha

1²+2²+3²+...+99²