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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2009}\right)\)
=\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2008}{2009}\)
=\(\frac{1}{2009}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)......\left(1-\frac{1}{100}\right).\)
<=> \(C=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{99}{100}\)
<=> \(C=\frac{1.2.3....99}{2.3.4....100}\)
<=> \(C=\frac{1}{100}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(1-1/2)(1-1/3)(1-1/4)...(1-1/2009)
=1/2*2/3*3/4*...*2008/2009
=\(\frac{1\cdot2\cdot3\cdot...\cdot2008}{2\cdot3\cdot4\cdot...\cdot2009}\)
=1/2009
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(=\frac{3}{2}.\frac{4}{3}....\frac{100}{99}=\frac{100}{2}=50\)
a) =3/2 . 4/3 . 5/4 ...100/99
=\(\frac{3.4.5...100}{2.3.4..99}\)
=\(\frac{100}{2}\)
b) =
![](https://rs.olm.vn/images/avt/0.png?1311)
Từ công thức:\(1+2+........+n=\frac{n.\left(n+1\right)}{2}\)
Cho \(n\in\)N*.CMR:\(\frac{1}{n}.\left(1+2+...+n\right)=\frac{n+1}{2}\)
Ta có:\(\frac{1}{n}.\left(1+2+......+n\right)=\frac{1}{n}.\frac{n\left(n+1\right)}{2}=\frac{n+1}{2}\)
Ta có:\(1+\frac{1}{2}\left(1+2\right)+......+\frac{1}{20}.\left(1+2+.....+20\right)\)
\(=1+\frac{1}{2}.\frac{2\left(2+1\right)}{2}+\frac{1}{3}.\frac{3.\left(3+1\right)}{2}+........+\frac{1}{20}.\frac{20\left(20+1\right)}{2}\)
\(=1+\frac{3}{2}+...............+\frac{21}{2}\)
\(=\frac{2+3+......+21}{2}\)
\(=\frac{230}{2}=165\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2.\(B=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.......\frac{49}{50}=\frac{1}{50}\)
tôi chỉ bn nè muốn làm thì hẳng hok thuộc đề bài vừa hok thuộc vùa nghĩ về bài sẽ nhưng thế nào
ai trả lời đi