CMR: \(\dfrac{1}{4}\)+\(\dfrac{1}{16}\) + \(\dfrac{1}{36}\) +....+ \(\dfrac{1}{10000}\) < \(_{\dfrac{1}{2}}\)
CMR : a, \(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+\dfrac{1}{100}+\dfrac{1}{144}+\dfrac{1}{196}+...+\dfrac{1}{10000}< \dfrac{1}{2}\)
b, \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< \dfrac{1}{3}\)
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)
Chứng minh rằng: \(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+...+\dfrac{1}{10000}< \dfrac{1}{2}\)
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)
\(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+...+\dfrac{1}{10000}=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)
\(=\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)< \dfrac{1}{4}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)=\dfrac{1}{4}\left(1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=\dfrac{1}{4}\left(1+1-\dfrac{1}{50}\right)=\dfrac{1}{4}\left(2-\dfrac{1}{50}\right)< \dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+...+\dfrac{1}{10000}< \dfrac{1}{2}\)
\(\left(\dfrac{1}{16}-1\right)\left(\dfrac{1}{25}-1\right)\left(\dfrac{1}{36}-1\right)...\left(\dfrac{1}{10000}-1\right)\)
Chứng minh rằng
S = \(\dfrac{1}{4}\)+\(\dfrac{1}{16}\)+\(\dfrac{1}{36}\)+.........+\(\dfrac{1}{10000}\)<\(\dfrac{1}{2}\)
S = \(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+......+\dfrac{1}{10000}\)
\(\Rightarrow S=\dfrac{1}{4.1}+\dfrac{1}{4.4}+\dfrac{1}{4.9}+.....+\dfrac{1}{4.2500}\)
\(\Rightarrow S=\dfrac{1}{4.\left(1+\dfrac{1}{4}+\dfrac{1}{9}+...+\dfrac{1}{2500}\right)}< \dfrac{1}{2}\)
\(\RightarrowĐPCM\)
(1-\(\dfrac{1}{4}\))x(1-\(\dfrac{1}{9}\))x(1-\(\dfrac{1}{16}\) )x...x(1-\(\dfrac{1}{10000}\))
\(\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right).....\left(1-\dfrac{1}{10000}\right)\)
\(=\dfrac{2^2-1}{2^2}\cdot\dfrac{3^2-1}{3^2}\cdot\dfrac{4^2-1}{4^2}\cdot\cdot\cdot\dfrac{100^2-1}{100^2}\)
\(=\dfrac{1.3.2.4.3.5.....99.101}{2.2.3.3.4.4....100.100}\)
\(=\dfrac{\left(1.2.3...99\right)}{2.3.4....100}\cdot\dfrac{3.4.5...101}{2.3.4....100}=\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{101}{200}\)
Ta có: \(\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)\cdot...\cdot\left(1-\dfrac{1}{10000}\right)\)
\(=\dfrac{-3}{4}\cdot\dfrac{-8}{9}\cdot\dfrac{-15}{16}\cdot...\cdot\dfrac{-9999}{10000}\)
\(=-\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{9999}{10000}\)
\(=\dfrac{-3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot1111\cdot9}{2^2\cdot3^2\cdot4^2\cdot...\cdot100^2}\)
\(=\dfrac{101}{200}\)
Chứng tỏ rằng: \(\dfrac{1}{2}< \dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{9801}+\dfrac{1}{10000}< \dfrac{37}{50}\)
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)
Tính tổng các phân số sau :
a) A = \(\dfrac{36}{1.3.5}+\dfrac{36}{3.5.7}+........+\dfrac{36}{45.47.49}\)
b) B= \(\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{9}\right).\left(1-\dfrac{1}{16}\right)......\left(1-\dfrac{1}{10000}\right)\)
c) C = 3+33+333+3333+...........+333........333
\(\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right).....\left(1-\dfrac{1}{10000}\right)\)
\(=\left(\dfrac{4}{4}-\dfrac{1}{4}\right)\left(\dfrac{9}{9}-\dfrac{1}{9}\right)\left(\dfrac{16}{16}-\dfrac{1}{16}\right)...\left(\dfrac{10000}{10000}-\dfrac{1}{10000}\right)\)
\(=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}....\cdot\dfrac{9999}{10000}\)
\(=\dfrac{3.8.15.....9999}{4.9.16.....10000}=\dfrac{\left(1.3\right)\left(2.4\right)\left(3.5\right)....\left(99.101\right)}{\left(2.2\right)\left(3.3\right)\left(4.4\right).....\left(100.100\right)}\)
\(=\dfrac{\left(1.2.3...99\right)\left(3.4.5....101\right)}{\left(2.3.4...100\right)\left(2.3.4...101\right)}=\dfrac{101.1}{100.2}=\dfrac{101}{200}\)
CMR C = \(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{9999}{10000}< \dfrac{1}{100}\)
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
Tính: \(E=\dfrac{\left(\dfrac{1}{2}-1\right).\left(\dfrac{1}{3}-1\right)...\left(\dfrac{1}{2002}-1\right).\left(\dfrac{1}{2003}-1\right)}{\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}...\dfrac{9999}{10000}}\)
Giải chi tiết giúp mình nha. Thanks
\(E=\dfrac{\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{2002}-1\right)\left(\dfrac{1}{2003}-1\right)}{\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot...\cdot\dfrac{9999}{10000}}\)
\(=\dfrac{\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2002}\right)\left(1-\dfrac{1}{2003}\right)}{\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{100^2}\right)}\)
\(=\dfrac{\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2002}\right)\left(1-\dfrac{1}{2003}\right)}{\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{100}\right)\left(1+\dfrac{1}{100}\right)}\)
\(=\dfrac{\dfrac{100}{101}\cdot\dfrac{101}{102}\cdot...\cdot\dfrac{2002}{2003}}{\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1+\dfrac{1}{100}\right)}\)
\(=\dfrac{100}{2003}:\left(\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\right)\)
\(=\dfrac{100}{2003}:\left(\dfrac{101}{2}\right)=\dfrac{100}{2003}\cdot\dfrac{2}{101}=\dfrac{200}{202303}\)