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

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)

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

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

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

tính nhé các bạn

chả có cái quy luật j cho mẫu số ak???

Bài này mình da làm roi dễ

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

\(\Rightarrow S=\dfrac{1}{4.1}+\dfrac{1}{4.4}+\dfrac{1}{4.9}+.....+\dfrac{1}{4.2500}\)

\(\Rightarrow S=\dfrac{1}{4.\left(1+\dfrac{1}{4}+\dfrac{1}{9}+...+\dfrac{1}{2500}\right)}< \dfrac{1}{2}\)

\(\RightarrowĐPCM\)

\(\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right).....\left(1-\dfrac{1}{10000}\right)\)

\(=\dfrac{2^2-1}{2^2}\cdot\dfrac{3^2-1}{3^2}\cdot\dfrac{4^2-1}{4^2}\cdot\cdot\cdot\dfrac{100^2-1}{100^2}\)

\(=\dfrac{1.3.2.4.3.5.....99.101}{2.2.3.3.4.4....100.100}\)

\(=\dfrac{\left(1.2.3...99\right)}{2.3.4....100}\cdot\dfrac{3.4.5...101}{2.3.4....100}=\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{101}{200}\)

Ta có: \(\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)\cdot...\cdot\left(1-\dfrac{1}{10000}\right)\)

\(=\dfrac{-3}{4}\cdot\dfrac{-8}{9}\cdot\dfrac{-15}{16}\cdot...\cdot\dfrac{-9999}{10000}\)

\(=-\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{9999}{10000}\)

\(=\dfrac{-3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot1111\cdot9}{2^2\cdot3^2\cdot4^2\cdot...\cdot100^2}\)

\(=\dfrac{101}{200}\)

Ta có : \(\dfrac{1}{4}\)= \(\dfrac{1}{2.2}\)> \(\dfrac{1}{2.3}\)

\(\dfrac{1}{9}\)= \(\dfrac{1}{3.3}\)> \(\dfrac{1}{3.4}\)

\(\dfrac{1}{16}\)=\(\dfrac{1}{4.4}\)> \(\dfrac{1}{4.5}\)

.......

\(\dfrac{1}{9801}\)= \(\dfrac{1}{99.99}\)> \(\dfrac{1}{99.100}\)

\(\dfrac{1}{10000}\)= \(\dfrac{1}{100.100}\)> \(\dfrac{1}{100.101}\)

\(\Rightarrow\) \(\dfrac{1}{4}\)+ \(\dfrac{1}{9}\)+ \(\dfrac{1}{16}\)+ ..... + \(\dfrac{1}{9801}\)+ \(\dfrac{1}{10000}\)> \(\dfrac{1}{2.3}\)+ \(\dfrac{1}{3.4}\)+ \(\dfrac{1}{4.5}\)+...+ \(\dfrac{1}{99.100}\)+\(\dfrac{1}{100.101}\)

= \(\dfrac{3-2}{2.3}\)+ \(\dfrac{4-3}{3.4}\)+ \(\dfrac{5-4}{4.5}\) +...+ \(\dfrac{100-99}{99.100}\)+ \(\dfrac{101-100}{100.101}\)

= \(\dfrac{3}{2.3}\)- \(\dfrac{2}{2.3}\) + \(\dfrac{4}{3.4}\)-\(\dfrac{3}{3.4}\)+ \(\dfrac{5}{4.5}\)-\(\dfrac{4}{4.5}\)+...+ \(\dfrac{100}{99.100}\)- \(\dfrac{99}{99.100}\)+ \(\dfrac{101}{100.101}\)-\(\dfrac{100}{100.101}\)

= \(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+....+ \(\dfrac{1}{99}\)-\(\dfrac{1}{100}\)+\(\dfrac{1}{100}\)-\(\dfrac{1}{101}\)

= \(\dfrac{1}{2}\)- \(\dfrac{1}{101}\) ; Mà \(\dfrac{1}{2}\)- \(\dfrac{1}{101}\)= \(\dfrac{99}{202}\)< \(\dfrac{1}{2}\)

\(\Rightarrow\) \(\dfrac{1}{2}\)< \(\dfrac{1}{4}\)+ \(\dfrac{1}{9}\)+ \(\dfrac{1}{16}\)+...+ \(\dfrac{1}{9801}\)+ \(\dfrac{1}{10000}\) (1)

\(=\left(\dfrac{4}{4}-\dfrac{1}{4}\right)\left(\dfrac{9}{9}-\dfrac{1}{9}\right)\left(\dfrac{16}{16}-\dfrac{1}{16}\right)...\left(\dfrac{10000}{10000}-\dfrac{1}{10000}\right)\)

\(=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}....\cdot\dfrac{9999}{10000}\)

\(=\dfrac{3.8.15.....9999}{4.9.16.....10000}=\dfrac{\left(1.3\right)\left(2.4\right)\left(3.5\right)....\left(99.101\right)}{\left(2.2\right)\left(3.3\right)\left(4.4\right).....\left(100.100\right)}\)

\(=\dfrac{\left(1.2.3...99\right)\left(3.4.5....101\right)}{\left(2.3.4...100\right)\left(2.3.4...101\right)}=\dfrac{101.1}{100.2}=\dfrac{101}{200}\)

ta có :
1/2 < 2/3
2/3 <3/4
.........
9999/10000 < 10000/10001
suy ra : A2 < 1/22/33/4*****9999/1000010000/10001
suy ra : A2 < 1/10001 < 1/10000= (1/100)2
suy ra A2 < (1/100)2 . Từ đó: A < 1/100

2 là mũ 2 nha bạn

\(E=\dfrac{\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{2002}-1\right)\left(\dfrac{1}{2003}-1\right)}{\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot...\cdot\dfrac{9999}{10000}}\)

\(=\dfrac{\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2002}\right)\left(1-\dfrac{1}{2003}\right)}{\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{100^2}\right)}\)

\(=\dfrac{\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2002}\right)\left(1-\dfrac{1}{2003}\right)}{\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{100}\right)\left(1+\dfrac{1}{100}\right)}\)

\(=\dfrac{\dfrac{100}{101}\cdot\dfrac{101}{102}\cdot...\cdot\dfrac{2002}{2003}}{\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1+\dfrac{1}{100}\right)}\)

\(=\dfrac{100}{2003}:\left(\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\right)\)

\(=\dfrac{100}{2003}:\left(\dfrac{101}{2}\right)=\dfrac{100}{2003}\cdot\dfrac{2}{101}=\dfrac{200}{202303}\)