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a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
b)\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{9900}=\frac{4949}{9900}\)
a)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
b)
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+....+\frac{1}{98\cdot99\cdot100}\)
\(=\frac{3-1}{1\cdot2\cdot3}+\frac{4-2}{2\cdot3\cdot4}+\frac{5-3}{3\cdot4\cdot4}+....+\frac{100-98}{98\cdot99\cdot100}\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+....+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1\cdot2}-\frac{1}{99\cdot100}\right)\)
\(=\frac{1}{2}\cdot\frac{4949}{9900}=\frac{4949}{19800}\)
Đặt A=(1-2/2*3)*(1-2/3*4)*(1-2*/4*5)*...*(1-2/99*100)
A=\(\frac{1\cdot4}{2\cdot3}\)*\(\frac{2\cdot5}{3\cdot4}\)*\(\frac{3\cdot6}{4\cdot5}\)*...*\(\frac{98\cdot101}{99\cdot100}\)
A=\(\frac{1\cdot101}{3\cdot99}\)
A=\(\frac{101}{297}\)
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}.\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
b) \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{35.37}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{35}-\frac{1}{37}\)
\(=\frac{1}{3}-\frac{1}{37}=\frac{34}{111}\)
c) \(\frac{7}{7.9}+\frac{7}{9.11}+\frac{7}{11.13}+...+\frac{7}{99.101}\)
\(=\frac{7}{2}.\left(\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{7}{2}.\left(\frac{1}{7}-\frac{1}{101}\right)=\frac{7}{2}\cdot\frac{94}{707}=\frac{47}{101}\)
a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow A< 1\)
b) \(B=\frac{1}{3}+\left(\frac{1}{3}\right)^2+...+\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrow3B=1+\frac{1}{3}+...+\left(\frac{1}{3}\right)^{99}\)
\(\Rightarrow3B-B=1-\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrow2B=1-\left(\frac{1}{3}\right)^{100}< 1\)
\(\Rightarrow2B< 1\)
\(\Rightarrow B< \frac{1}{2}\)
Đặt B = \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
Đặt C = \(\frac{1}{51.100}+\frac{1}{52.99}+...+\frac{1}{75.76}\)(sửa lại đề)
=> 151C = \(\frac{151}{51.100}+\frac{151}{52.99}+...+\frac{151}{75.76}\)
=> 151C =\(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)
=> C = \(\frac{\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}}{151}\)
Khi A = B : C
= \(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right):\left(\frac{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}{151}\right)=151\)
Vậy A = 151
\(\left(1-\frac{2}{2\times3}\right)\times\left(1-\frac{2}{3\times4}\right)\times\left(1-\frac{2}{4\times5}\right)\times...\times\left(1-\frac{2}{99\times100}\right)\)
=\(\frac{2}{2}-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+\frac{2}{4}-\frac{2}{5}+...+\frac{2}{99}-\frac{2}{100}\)
=\(\frac{2}{2}-\frac{2}{100}\)
=\(\frac{98}{100}\)
=\(\frac{49}{50}\)
A= 1-\(\frac{1}{2}\) +\(\frac{1}{3}\) - \(\frac{1}{4}\) +\(\frac{1}{5}\)- \(\frac{1}{6}\) + ...+ \(\frac{1}{99}\) - \(\frac{1}{100}\)
= 1+ \(\frac{1}{2}\) + \(\frac{1}{3}\) + \(\frac{1}{4}\) + \(\frac{1}{5}\) + \(\frac{1}{6}\) + ...+ \(\frac{1}{99}\) + \(\frac{1}{100}\) - 2 ( \(\frac{1}{2}\) + \(\frac{1}{4}\) + \(\frac{1}{6}\) + ...+ \(\frac{1}{100}\) )
= 1+ \(\frac{1}{2}\) + \(\frac{1}{3}\) + \(\frac{1}{4}\) + ...+ \(\frac{1}{99}\) + \(\frac{1}{100}\)
= \(\frac{1}{51}\) + \(\frac{1}{52}\) +...+ \(\frac{1}{100}\)
= (\(\frac{1}{51}\) + \(\frac{1}{52}\) + ... + \(\frac{1}{75}\) ) + ( \(\frac{1}{76}\) + \(\frac{1}{77}\) + ... + \(\frac{1}{100}\) )
Ta có : \(\frac{1}{51}\) > \(\frac{1}{52}\) > \(\frac{1}{53}\) > ... > \(\frac{1}{75}\)
\(\frac{1}{76}\) > \(\frac{1}{77}\) > \(\frac{1}{78}\) > ... > \(\frac{1}{100}\)
=> A > \(\frac{1}{75}.25\) + \(\frac{1}{100}.25\) = \(\frac{1}{3}\) + \(\frac{1}{4}\) = \(\frac{7}{12}\)
=> A< \(\frac{1}{51}.25\) + \(\frac{1}{75}.25\) < \(\frac{1}{50}.25\) + \(\frac{1}{75}.25\) = \(\frac{1}{2}\) + \(\frac{1}{3}\) = \(\frac{5}{6}\)
Vậy \(\frac{7}{12}\) < A < \(\frac{5}{6}\)
Tick nha
đề bài là j ý ạ
\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{49}{100}\)