4A = 4.[1.2.3 + 2.3.4 + 3.4.5 + … + (n – 1).n.(n + 1)]

4A = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + … + (n – 1).n.(n + 1).4

4A = 1.2.3.4 + 2.3.4.(5 – 1) + 3.4.5.(6 – 2) + … + (n – 1).n.(n + 1).[(n + 2) – (n – 2)]

4A = 1.2.3.4 + 2.3.4.5 – 1.2.3.4 + 3.4.5.6 – 2.3.4.5 + … + (n – 1).n(n + 1).(n + 2) – (n – 2).(n – 1).n.(n + 1)

4A = (n – 1).n(n + 1).(n + 2)

A = (n – 1).n(n + 1).(n + 2) : 4.

cau a thi sao ha ban ? 

ok thanks ban nhe

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

549 + X = 1326
X = 1326 - 549
X = 777
X - 636 = 5618
X = 5618 + 636
X = 6254

549 ,1326 ở đâu zậy bạn  !!! :/

Đặt A = 1.2.3 + 2.3.4 + 3.4.5 + ... + 28.29.30

4A = 1.2.3.(4-0) + 2.3.4.(5-1) + 3.4.5.(6-2) + ... + 28.29.30.(31-27)

4A = 1.2.3.4 - 0.1.2.3. + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + ... + 28.29.30.31 - 27.28.29.30

4A = 28.29.30.31 - 0.1.2.3

4A = 28.29.30.31

\(A=\frac{28.29.30.31}{4}=7.29.30.31=188790\)

Theo cách tính trên ta dễ dàng tính được:

1.2.3 + 2.3.4 + 3.4.5 + ... + (n - 1).n.(n + 1) = \(\frac{\left(n-1\right).n.\left(n+1\right).\left(n+2\right)}{4}\)

4(1.2.3) = 1.2.3.4 - 0.1.2.3

4(2.3.4) = 2.3.4.5 - 1.2.3.4

4(3.4.5) = 3.4.5.6 - 2.3.4.5

....................................

4(n-1)n(n+1) = (n-1)n(n+1)(n+2) - (n-2)(n-1)n(n+1)

=> 4 B = (n-1)n(n+1)(n+2) => B= (n-1)n(n+1)(n+2):4

4(1.2.3)=1.2.3.4 - 0.1.2.3

4(2.3.4)=2.3.4.5 - 1.2.3.4

4(3.4.5)=3.4.5.6 - 2.3.4.5

.........................

......................................

.......................................

thank

=\(\frac{\left(n-1\right)n\left(n+1\right)\left(n+2\right)}{4}\)

4B = 1.2.3.4 + 2.3.4.4 + ... + (n-1)n(n+1).4

= 1.2.3.4 - 0.1.2.3 + 2.3.4.5 - 1.2.3.4 + ... + (n-1)n(n+1)(n+2) - [(n-2)(n-1)n(n+1)]

= (n-1)n(n+1)(n+2) - 0.1.2.3

= (n-1)n(n+1)(n+2)

suy ra \(B = {(n-1)n(n+1)(n+2)\over 4}\)

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

B=1.2.3+2.3.4+3.4.5+...+n.(n+1).(n+2)

4B=4.(1.2.3+2.3.4+3.4.5+...+n.(n+1).(n+2)

4B=1.2.3.4+2.3.4.4+3.4.5.4+...+n.(n+1).(n+2).4

4B=1.2.3.4+2.3.4.(5-1)+3.4.5.(6-2)+...+n.(n+1)+(n+2).(n+3-4)

4B=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+...+n.(n+1).(n+2).(n+3)-(n-1).n.(n+1).(n+2)

4B=(1.2.3.4-1.2.3.4)+(2.3.4.5-2.3.4.5)+...+[(n-1).n.(n+1).(n+2)-(n-1).n.(n+1).(n+2)]+n.(n+2).(n+2).(n+3)

4B=0+0+...+0+n.(n+1).(n+2).(n+3)

4B=n.(n+1).(n+2).(n+3)

B=[n.(n+1).(n+2).(n+3)]:4

B=1.2.3+2.3.4+3.4.5+...+n(n+1)(n+2)

  ={1.2.3.(4-0)+2.3.4(5-1)+3.4.5.(6-2)+...+n(n+1)(n+2)[(n+3)-(n-1)]} : 4

  = [1.2.3.4+2.3.4.5+3.4.5.6+...+n(n+1)(n+2)(n+3) - 1.2.3.4 - 2.3.4.5 - 3.4.5.6 - ... - n(n+1)(n+2)(n-1)] : 4

  =\(\frac{\text{ n(n+1)(n+2)(n+3) }}{4}\)

B = \(\frac{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}{4}\)