Đặt A= 1.2.3+2.3.4+3.4.5+...+101.102.103

=>4A=1.2.3.4+2.3.4.4+3.4.5.4+...+101.102.103.4

=1.2.3.(4-0)+2.3.4.(5-1)+3.4.5.(6-2)+...+101.102.103.(104-100)

=1.2.3.4-0.1.2.3+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+...+101.102.103.104-100.101.102.103

=101.102.103.104-0.1.2.3

=110355024

=>A=110355024:4=27588756

Lời giải:

Gọi tổng trên là A

$2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{101.102.103}$

$=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{103-101}{101.102.103}$

$=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{101.102}-\frac{1}{102.103}$

$=\frac{1}{1.2}-\frac{1}{102.103}=\frac{2626}{5253}$

$\Rightarrow A=\frac{1313}{5253}$

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

\(B=\dfrac{1}{1.2.3}+\dfrac{1}{3.4.5}+...+\dfrac{1}{8.9.10}\)

\(B=2.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(B=2.\left(1-\dfrac{1}{10}\right)\)

\(B=2.\dfrac{9}{10}\)

\(B=\dfrac{9}{5}\)

anh ơi , đại học rồi mà ko giải đc bài này ạ?

hmm

2P=2/1.2.3+2/2.3.4+2/3.4.5+2/10.11.12
2P=1/1.2-1/2.3+1/2.3-1/3.4+1/3.4-1/4.5+.....+1/10.11-1/11.12
2P=1/1.2-1/11.12
2P=1/2-1/132
2P=66/132-1/132
2P=65/132
 P=65/264

\(P=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{10.11.12}\)

\(P=\dfrac{1}{2}-\dfrac{1}{11.12}\)

\(P=\dfrac{65}{132}\)

\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}...+\frac{2}{8.9.10}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{90}\right)=\frac{11}{45}\)

     \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{90}\right)\)

\(=\frac{1}{2}.\frac{22}{45}\)

\(=\frac{11}{45}\)

=(1/1-1/2-1/3)+(1/2-1/3-1/4)+...+(1/18-1/19-1/20)

=1/1-1/20=19/20

k nha

dung luon nhung van tat qua