s=1/1*2-1/2*3+1/2*3-1/3*4+....+1/2009*2010-1/210*2011

 =1/1*2-1/2010*2011

<1/1*2

\(S=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{2009\cdot2010\cdot2011}\)

\(S=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{2009\cdot2010}-\frac{1}{2010\cdot2011}\)

\(S=\frac{1}{1\cdot2}-\frac{1}{2010\cdot2011}\)

\(S=\frac{1}{2}-\frac{1}{2010\cdot2011}< \frac{1}{2}\)

=> S < P

Ta có :

\(S=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...............+\dfrac{2}{2009.2010.2011}\)

\(S=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+.........+\dfrac{1}{2009.2010}-\dfrac{1}{2010.2011}\)

\(S=\dfrac{1}{1.2}-\dfrac{1}{2010.2011}\)

\(S=\dfrac{1}{2}-\dfrac{1}{4042110}\) \(< \dfrac{1}{2}\)

\(\Rightarrow S< Q\)

S=\(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}\)+......+\(\frac{2}{2009.2010.2011}\)

S=\(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+.......+\frac{1}{2009.2010}-\frac{1}{2010.2011}\)

S=\(\frac{1}{2}-\frac{1}{2010.2011}<\frac{1}{2}\)

Suy ra: S<P

\(S=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{2009.2010.2011}\)

\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{2009.2010}-\dfrac{1}{2010.2011}\)

\(=\dfrac{1}{1.2}-\dfrac{1}{2010.2011}\)

\(=\dfrac{1}{2}-\dfrac{1}{4042110}< \dfrac{1}{2}\)

\(\Rightarrow\) \(S< P\)

Vậy \(S< P\)

Cảm ơn nhá

1/2-1/2010.2011

cậu có biết tách ko?

Bài 1 :

\(S=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2009.2010.2011}\)

    \(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2009.2010}-\frac{1}{2010.2011}\)

    \(=\frac{1}{1.2}-\frac{1}{2010.2011}<\frac{1}{2}\)

Vậy \(S<\frac{1}{2}\)

Bài 2:

Làm nhiều rồi vào trong chỗ góc học tập của tớ mà coi 

THANKS BN NHÌU

Tổng quát: \(\frac{2}{\left(a-1\right)a\left(a+1\right)}=\frac{1}{\left(a-1\right).a}-\frac{1}{a\left(a+1\right)}\)

Ta có: \(S=\frac{2}{1.2.3}+\frac{2}{2.3.4}+.....+\frac{2}{2013.2014.2015}\)

\(S=\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+.....+\left(\frac{1}{2013.2014}-\frac{1}{2014.2015}\right)\)

\(S=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{2013.2014}-\frac{1}{2014.2015}\)

\(S=\frac{1}{1.2}-\frac{1}{2014.2015}=\frac{1}{2}-\frac{1}{2014.2015}<\frac{1}{2}\)

Vậy....................

S=(2/1.2-2/2.3)+(2/2.3-2/3.4)+(2/3.4-2/4.5)+...........+(2/2013.2014-2/2014-2/2015)

S=(2/1.2-2/2014.2015):2

S=1-2/2014.2/2015

--> S>1/2

Ta có: \(\frac{3n+2}{n\left(n+1\right)\left(n+2\right)}\)

\(=\frac{2n+2+n}{n\left(n+1\right)\left(n+2\right)}=\frac{2\left(n+1\right)}{n\left(n+1\right)\left(n+2\right)}+\frac{n}{n\left(n+1\right)\left(n+2\right)}\)

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

\(=\frac{1}{n}+\frac{1}{n+1}-\frac{2}{n+2}\)

Do đó, ta có: \(\frac{5}{1\cdot2\cdot3}=\frac{3\cdot1+2}{1\cdot2\cdot3}=\frac11+\frac{1}{1+1}-\frac{2}{1+2}=1+\frac12-\frac23\)

\(\frac{8}{2\cdot3\cdot4}=\frac{3\cdot2+2}{2\cdot3\cdot4}=\frac12+\frac13-\frac24\)

...

Do đó, ta có: \(S=1+\frac12-\frac23+\frac12+\frac13-\frac24+\frac13+\frac14-\frac25+\ldots+\frac{1}{n}+\frac{1}{n+1}-\frac{2}{n+2}\)

\(=1+\left(\frac12+\frac12\right)+\left(-\frac23+\frac13+\frac13\right)+\left(-\frac24+\frac14+\frac14\right)+\cdots+\left(-\frac{2}{n}+\frac{1}{n}+\frac{1}{n}\right)-\frac{2}{n+1}+\frac{1}{n+1}-\frac{2}{n+2}\)

\(=1+1-\frac{1}{n+1}-\frac{2}{n+2}<2\)

=>\(S_{2022}=\frac{5}{1\cdot2\cdot3}+\frac{8}{2\cdot3\cdot4}+\cdots+\frac{6068}{2022\cdot2023\cdot2024}<2\)