A  = 1.2.3 + 2.3.4 + ....+ 48.49.50

=> 4A = 1.2.3.4 + 2.3.4.(5-1) + ...+ 48.49.50.(51-17)

= 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + .....+ 48.49.50.51 - 47.48.49.50

= 48.49.50.51

=> A =  48.49.50.51:4 = 12.49.50.51

bài b) làm tương tự nha

\(A=1.2.3+2.3.4+3.4.5+...+48.49.50\)

\(4A=1.2.3.4+2.3.4.4+3.4.5.4+...+48.49.50.4\)

\(4A=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)+...+48.49.50.\left(51-47\right)\)

\(4A=1.2.3.4-0.1.2.3+2.3.4.5-1.2.3.4+...+48.49.50.51-48.48.49.50\)

\(4A=48.49.50.51\)

\(A=\dfrac{48.49.50.51}{4}=1499400\)

A=1499400 nhe ban !

4F=4.[1.2.3 + 2.3.4 + 3.4.5 + 4.5.6 + . . . . . . + 48.49.50] 
4F=1.2.3.4 +2.3.4.4 +3.4.5.4 +4.5.6.4 +.........+48.49.50.4 
4F=1.2.3.4 +2.3.4.(5-1) + 3.4.5.(6-2) +4.5.6(7-3)+....+ 48.49.50(51-47) 
4F=1.2.3.4 +2.3.4.5 --1.2.3.4 + 3.4.5.6--2.3.4.5 + 4.5.6.7-3.4.5.6+....+ 48.49.50.51--47.48.49.50 
4F =48.49.50.51 
F=(48.49.50.51)/4 

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

đặt A = 1.2.3 + 2.3.4 + 3.4.5 + ... + 48.49.50

4A = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + ...+ 48.49.50.4

4A = 1.2.3.4 + 2.3.4.(5-1) + 3.4.5.(6-2) + ... + 48.49.50.(51-47)

4A = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + ... + 48.49.50.51 - 47.48.49.50

4A = 48.49.50.51

A = \(\frac{48.49.50.51}{4}\)

A = 1499400

Đặt A=1.2.3+2.3.4+3.4.5+........+48.49.50

     4A=1.2.3.4+2.3.4.4+..........+48.49.50.4

         =1.2.3.4+2.3.4.(5-1)+.........+48.49.50.(51-47)

         =1.2.3.4+2.3.4.5-1.2.3.4+...........+48.49.50.51-47.48.49.50

         =48.49.50.51

         =5997600

       A=1499400

Vậy A=1499400

Đặt \(A=1\cdot2\cdot3+2\cdot3\cdot4+........+48\cdot49\cdot50\)

\(\Rightarrow4A=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot4+......+48\cdot49\cdot50\cdot4\)

\(=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot\left(5-1\right)+..........+48\cdot49\cdot50\cdot\left(51-47\right)\)

\(=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot5-1\cdot2\cdot3\cdot4+......+48\cdot49\cdot50\cdot51-47\cdot48\cdot49\cdot50\)

\(=48\cdot49\cdot50\cdot51\)

\(\Rightarrow A=\frac{48\cdot49\cdot50\cdot51}{4}\)

\(C=1.2.3+2.3.4+...+48.49.50\)

\(\Rightarrow4C=1.2.3.4+2.3.4.4+...+48.49.50.4\)

\(=1.2.3.4+2.3.4.\left(5-1\right)+...+48.49.50.\left(51-47\right)\)

\(=1.2.3.4+2.3.4.5-1.2.3.4+...+48.49.50.51-47.48.49.50\)

\(=48.49.50.51\)

\(\Rightarrow C=\frac{48.49.50.51}{4}=1499400\)