Tính:
.\(\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}}\)
Những câu hỏi liên quan
a) Chứng tỏ rằng \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)
b) Đặt A = \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2013+2014}\); Đặt B = \(\frac{1}{1008.2014}+\frac{1}{1009.2013}+...+\frac{1}{2014.1008}\)
Chứng tỏ rằng \(\frac{A}{B}\)là số nguyên
a) Chứng tỏ rằng
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)
b) Đặt A = \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\); B = \(\frac{1}{1008.2014}+\frac{1}{1009.2013}+...+\frac{1}{2014.1008}\)
Chứng tỏ rằng \(\frac{A}{B}\)là số nguyên
\(\frac{\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{199.200}}{\frac{1}{101}+\frac{1}{102}+...\frac{1}{200}}=1\)
Hãy chứng minh
Đặt \(A=\frac{\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{199.200}}{\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}}\)
Tử số của A = \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{199.200}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
\(\Rightarrow A=1\left(đpcm\right)\)
Đúng 1
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Cho A = \frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+.....+\frac{1}{199.200}1.21+3.41+5.61+.....+199.2001
B= \frac{1}{101.200}+\frac{1}{102.199}+.....+\frac{1}{200.101}101.2001+102.1991+.....+200.1011
Tính A:B
Cho A = \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+.....+\frac{1}{199.200}\)
B= \(\frac{1}{101.200}+\frac{1}{102.199}+.....+\frac{1}{200.101}\)
Tính A:B
Cho :
A=\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}\)
B=\(\frac{1}{101.200}+\frac{1}{102.199}+...+\frac{1}{199.102}\)
Hãy tính \(\frac{A}{B}\)
Chứng minh rằng:
a)\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}< \frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
b)\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}< 1-\frac{1}{2.3}\)
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\(B=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(B=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(B=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(B< \frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\)
\(B< \frac{50}{60}\Leftrightarrow B< \frac{5}{6}\)
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{200}=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)
$\frac{1}{101}$$+$$\frac{1}{102}$$+$$\frac{1}{103}$$+$ $.............$ $+$$\frac{1}{200}$
