S=1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 +...+ 1/2010.2011 - 1/2011.2012

S=1/1.2 - 1/2011.2012<1/2

=>S<P

75:x=3(du 3 )

\(S=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+........+\frac{2}{2010.2011.2012}\)

\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+......+\frac{1}{2010.2011}-\frac{1}{2011.2012}\)

\(=\frac{1}{1.2}-\frac{1}{2011.2012}\)

\(=\frac{1}{2}-\frac{1}{2011.2012}\)

mà \(\frac{1}{2}-\frac{1}{2011.2012}< \frac{1}{2}\)

\(\Rightarrow S< P\)

\(M=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{10.11}-\frac{1}{11.12}\)

\(=\frac{1}{2}-\frac{1}{11.12}\)

 \(=\frac{65}{132}\)

65/132

65/132

2A=2(1/1.2.3+1/2.3.4+...+1/98.99.100)

2A=1/1.2-1/2.3+1/2.3-1/3.4+1/3.4-...+1/98.99-1/99.100

2A=1/1.2-1/99.100

2A=4949/9900

A=4949/9900:2

A=4949/19800

                                         Vậy A=4949/198000

Lưu Trường An làm đúng rồi

Giải:

Ta có nhận xét:

\(\frac{1}{1.2}-\frac{1}{2.3}=\frac{3-1}{1.2.3}=\frac{2}{1.2.3}\)

\(\frac{1}{2.3}-\frac{1}{3.4}=\frac{4-2}{2.3.4}=\frac{2}{2.3.4}\)

=>\(\frac{1}{1.2.3}=\frac{1}{3}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)\)

\(\frac{1}{2.3.4}=\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)\)

Do đó M=\(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{10.11}-\frac{1}{11.12}\right)\)

=\(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{11.12}\right)=\frac{1}{2}-\frac{1}{11.12}\)

=\(\frac{1}{2}.\frac{65}{132}=\frac{65}{124}\)

Vậy M=65/124

M=\(\frac{65}{124}\)

65/124 nha bạn

Ta có : \(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)

\(\Leftrightarrow2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{18.19.20}\)

\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{18.19}-\frac{1}{19.20}\)

\(=\frac{1}{2}-\frac{1}{19.20}=\frac{189}{380}\)

\(\Rightarrow B=\frac{189}{760}\)

\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}\left(\frac{1}{18.19}-\frac{1}{19.20}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{19.20}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{380}\right)\)

\(=\frac{1}{2}.\frac{189}{380}=\frac{189}{760}\)

\(B=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{18.19.20}\right).\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right).\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{380}\right)\)

\(=\frac{179}{760}\)

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

=1+\(\frac{1}{101}\)

=\(\frac{102}{101}\)

1/1.2.3 = 1/2 .[1/1.2 - 1 / 2.3]

1/2.3.4 = 1/2[ 1/2- 1/3 ] 

...................

1/99.100.101 = 1/2[ 1/99. 100 - 1/100.101]

=> A= 1/2 [ 1/1.2- 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/ 4.5 +.........+ 1/99 .100 - 1/100. 101]

A = 1/2 . [1/1.2 -1/100 .101]

A= 1/2 . 5049 /10100 = 5049 / 20200.

Mình nghĩ là vậy đó.

a=100/101

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{98.99.100}=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(\Leftrightarrow\frac{1}{k}=\frac{1}{2}\Rightarrow k=2\)

\(A=2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\right)\)

\(A=2.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)

\(A=2.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(A=2\cdot\frac{4949}{9900}=\frac{4949}{4950}\)