c = 1.2.3+2.3.4+3.4.5+...+48.49.50 = ?
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S
17 tháng 3 2017
\(A=1.2.3+2.3.4+3.4.5+...+48.49.50\)
\(4A=1.2.3.4+2.3.4.4+3.4.5.4+...+48.49.50.4\)
\(4A=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)+...+48.49.50.\left(51-47\right)\)
\(4A=1.2.3.4-0.1.2.3+2.3.4.5-1.2.3.4+...+48.49.50.51-48.48.49.50\)
\(4A=48.49.50.51\)
\(A=\dfrac{48.49.50.51}{4}=1499400\)
20 tháng 9 2015
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
KM
1
17 tháng 7 2019
\(C=1.2.3+2.3.4+...+48.49.50\)
\(\Rightarrow4C=1.2.3.4+2.3.4.4+...+48.49.50.4\)
\(=1.2.3.4+2.3.4.\left(5-1\right)+...+48.49.50.\left(51-47\right)\)
\(=1.2.3.4+2.3.4.5-1.2.3.4+...+48.49.50.51-47.48.49.50\)
\(=48.49.50.51\)
\(\Rightarrow C=\frac{48.49.50.51}{4}=1499400\)
LB
7 tháng 5 2018
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
8 tháng 5 2018
\(P=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)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
NT
1
9 tháng 6 2017
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
VH
2 tháng 11 2019
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
BA
2 tháng 11 2019
Đặt \(A=1\cdot2\cdot3+2\cdot3\cdot4+........+48\cdot49\cdot50\)
\(\Rightarrow4A=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot4+......+48\cdot49\cdot50\cdot4\)
\(=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot\left(5-1\right)+..........+48\cdot49\cdot50\cdot\left(51-47\right)\)
\(=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot5-1\cdot2\cdot3\cdot4+......+48\cdot49\cdot50\cdot51-47\cdot48\cdot49\cdot50\)
\(=48\cdot49\cdot50\cdot51\)
\(\Rightarrow A=\frac{48\cdot49\cdot50\cdot51}{4}\)
15 tháng 8 2018
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
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