Ta có \(k\left(k+1\right)\left(k+2\right)=\dfrac{1}{4}k\left(k+1\right)\left(k+2\right)\cdot4\)

\(=\dfrac{1}{4}k\left(k+1\right)\left(k+2\right)\left[\left(k+3\right)-\left(k-1\right)\right]\\ =\dfrac{1}{4}k\left(k+1\right)\left(k+2\right)\left(k+3\right)-\dfrac{1}{4}\left(k-1\right)k\left(k+1\right)\left(k+2\right)\)

Từ đó ta được \(S=\dfrac{1}{4}\cdot1\cdot2\cdot3\cdot4-\dfrac{1}{4}\cdot0\cdot1\cdot2\cdot3+...+\dfrac{1}{4}\cdot9\cdot10\cdot11\cdot12-\dfrac{1}{4}\cdot8\cdot9\cdot10\cdot11\\ \Leftrightarrow S=\dfrac{1}{4}\cdot9\cdot10\cdot11\cdot12\\ \Leftrightarrow4S+1=9\cdot10\cdot11\cdot12+1=11881=109^2\left(đpcm\right)\)

Ở sbt 6 tập mấy ko nhớ có bài tương tự trong ngoặc, mở phần lời giải ra để tính trong ngoặc nha

\(S=1.2.3+2.3.4+3.4.5+...+9.10.11\)

\(4S=1.2.3.4+2.3.4.\left(5-1\right)+3.4.5.\left(6-2\right)+...+9.10.11.\left(12-8\right)\)

\(=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+...+9.10.11.12-8.9.10.11\)

\(=9.10.11.12\)

\(4S+1=9.10.11.12+1=\left(9.12\right).\left(10.11\right)+1=108.110+1\)

\(=\left(109-1\right)\left(109+1\right)+1=109^2-1+1=109^2\)

Ta có đpcm. 

Ta có \(k\left(k+1\right)\left(k+2\right)=\dfrac{1}{4}k\left(k+1\right)\left(k+2\right)\cdot4\)

\(=\dfrac{1}{4}k\left(k+1\right)\left(k+2\right)\left[\left(k+3\right)-\left(k-1\right)\right]\\ =\dfrac{1}{4}k\left(k+1\right)\left(k+2\right)\left(k+3\right)-\dfrac{1}{4}\left(k-1\right)k\left(k+1\right)\left(k+2\right)\)

Từ đó ta được \(S=\dfrac{1}{4}\cdot1\cdot2\cdot3\cdot4-\dfrac{1}{4}\cdot0\cdot1\cdot2\cdot3+...+\dfrac{1}{4}\cdot9\cdot10\cdot11\cdot12-\dfrac{1}{4}\cdot8\cdot9\cdot10\cdot11\\ \Leftrightarrow S=\dfrac{1}{4}\cdot9\cdot10\cdot11\cdot12\\ \Leftrightarrow4S+1=9\cdot10\cdot11\cdot12+1=11881=109^2\left(đpcm\right)\)

2Q=\(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+.........+\frac{1}{9.10}-\frac{1}{10.11}\)

2Q=\(\frac{1}{1.2}-\frac{1}{10.11}\)

2Q=\(\frac{1}{2}-\frac{1}{110}\)

2Q=\(\frac{55}{110}-\frac{1}{110}\)

2Q=\(\frac{54}{110}\)

Q=\(\frac{54}{110}:2\)

Q=\(\frac{27}{110}\)

bằng 27/100

bằng 27/110

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 

3F= 1.2.(3-0)+ 2.3.(4-1)+...+ n.(n+1).[(n+2)-(n-1)] 
=[1.2.3+ 2.3.4+...+ (n-1)n(n+1)+ n(n+1)(n+2)]- [0.1.2+ 1.2.3+...+(n-1)n(n+1)] 
=n(n+1)(n+2) 
=>F 

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

=> 4H=1.2.3(4-0)+2.3.4(5-1)+...+n(n+1)(n+2)((n+3)-(n-1))

=1.2.3.4-0.1.2.3+2.3.4.5-1.2.3.4+...+n(n+1)(n+2)(n+3)-(n-1).n(n+1)(n+2)

=n(n+1)(n+2)(n+3)

Nhân biểu thức S với số 5, ta có:

5.S = 1.2.3.4.5 + 2.3.4.5.5 + 3.4.5.6.5 + ... + 97.98.99.100.5

Biểu diễn số 5 ở mỗi số hạng vế phải bằng phép trừ thích hợp: 5 = 5 - 0 = 6 - 1 = 7 - 2 = ... = 101 - 96, ta có

5.S = 1.2.3.4.(5 - 0) + 2.3.4.5.(6 - 1) + 3.4.5.6.(7 - 2) + ...+ 97.98.99.100.(101 - 96)

     = (1.2.3.4.5 - 1.2.3.4.0) + (2.3.4.5.6 - 2.3.4.5.1) + (3.4.5.6.7 - 3.4.5.6.2) + ... + (97.98.99.100.101 - 97.98.99.100.96)

     = 1.2.3.4.5 - 0.1.2.3.4 + 2.3.4.5.6 - 1.2.3.4.5 + 3.4.5.6.7 - 2.3.4.5.6 + ... + 97.98.99.100.101 - 96.97.98.99.100

     = 97.98.99.100.101 - 0.1.2.3.4

     = 97.98.99.100.101

Suy ra  

  S = 97.98.99.100.101/5 = 97.98.99.20.101. Đến đây thì bạn dùng máy tính bấm ra S=1901009880

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