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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 :
\(S=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...............+\dfrac{2}{2009.2010.2011}\)
\(S=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+.........+\dfrac{1}{2009.2010}-\dfrac{1}{2010.2011}\)
\(S=\dfrac{1}{1.2}-\dfrac{1}{2010.2011}\)
\(S=\dfrac{1}{2}-\dfrac{1}{4042110}\) \(< \dfrac{1}{2}\)
\(\Rightarrow S< Q\)
\(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2009.2010.2011}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2009.2010}-\frac{1}{2010.2011}\)
\(=\frac{1}{2}-\frac{1}{2010.2011}< \frac{1}{2}\)
Vậy...
a/
\(b=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)
\(2b=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{99-97}{97.99}=\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}=\)
\(=1-\dfrac{1}{99}=\dfrac{98}{99}\Rightarrow b=\dfrac{98}{2.99}=\dfrac{49}{99}\)
b/
\(c=\dfrac{3-1}{1.2.3}+\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}=\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+\dfrac{1}{98.99}-\dfrac{1}{99.100}=\)
\(=\dfrac{1}{2}-\dfrac{1}{99.100}\)
c/
\(\dfrac{2}{5}.d=\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}+\dfrac{101-99}{99.100.101}=\)
\(=\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}+\dfrac{1}{99.100}-\dfrac{1}{100.101}=\)
\(=\dfrac{1}{2.3}-\dfrac{1}{100.101}\Rightarrow d=\left(\dfrac{1}{2.3}-\dfrac{1}{100.101}\right):\dfrac{2}{5}\)
đặt S=1.2.3+2.3.4+....+47.48.49
4S=1.2.3.(4-0)+2.3.4.(5-1)+...+47.48.49.(50-46)
4S=1.2.3.4-1.2.3+2.3.4.5-1.2.3.4+....+47.48.49.50-46.47.48.49
4S=47.48.49.50-1.2.3
S=(47.48.49.50-1.2.3):4
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
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
tao có:
2p=2/1.2.3+2/2.3.4+...+2/n.n(+1)n(n+2)
2p=3-1/1.2.3+4-2/1.2.3+...+(n+2)-n/n.(n+1).(n+2)
2p=3/1.2.3-1/1.2.3+4/2.3.4-2/2.3.4+...+(n+2)/n.(n+1).(n+2)-n/n.(n+1).(n+2)
2p=1/1.2-1/2.3+1/2.3-1/3.4+...+1/n.(n+1)-1/(n+1).(n+2)
2p=1/1.2-1/(n+1).(n+2)
2p=(n+!).(n+2)-2/(2n+2).(n+2)
suy ra p=(n+1).(n+2)-2/(2n+2).(2n+4)
2s=3-1/1.2.3+4-2/1.2.3+...+50-48/48.49.50
2s=3/1.2.3-1/1.2.3+4/2.3.4-2/2.3.4+...+50/49.50.48-48/48.50.49
2s=1/1.2-1/2.3+1/2.3-1/3.4+...+1/48.49-1/49.50
2s=1/1.2-1/49.50
'2s=1/2-1/2450
2s=1225/2450-1/2450
2s=1224/2450
s=612/1225
\(P=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)1
\(P=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\right)\)
\(P=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(P=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(P=\frac{\left(\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)}{2}\)
S cx tinh giong v
\(S=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2009.2010.2011}\)
\(S=2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2009.2010.2011}\right)\)
\(S=2.\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+...+\frac{1}{2009.2010}-\frac{1}{2010.2011}\right)\)
\(S=1.\left(\frac{1}{1.2}-\frac{1}{2010.2011}\right)\)
\(S=\frac{1}{1.2}-\frac{1}{2010.2011}\)
\(S=\frac{1}{2}-\frac{1}{2010.2011}< \frac{1}{2}\)
Vậy \(S< \frac{1}{2}\)
Chúc bạn học tốt !!!
Áp dụng công thức :
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{6}\right)=\frac{1}{2}.\frac{2}{6}=\frac{1}{6}=\frac{1}{1.2.3}\)
Chúc bạn học tốt !!!