vì bạn có l.i.k.e đâu mà trả lời

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

1.2.3 = 1/4 . (1.2.3.4 - 0.1.2.3)

2.3.4 = 1/4 . (2.3.4.5 - 1.2.3.4)

3.4.5 = 1/4 . (3.4.5.6 - 2.3.4.5)

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

99.100.101 = 1/4 . (99.100.101.102 - 98.99.100.101)

C = 1.2.3+2.3.4+3.4.5+.........+99.100.101

C= 1/4 . (99.100.101.102 - 98.99.100.101)

CHUC BN HOK GIỎI!

25497450

a/

\(b=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)

\(2b=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{99-97}{97.99}=\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}=\)

\(=1-\dfrac{1}{99}=\dfrac{98}{99}\Rightarrow b=\dfrac{98}{2.99}=\dfrac{49}{99}\)

b/

\(c=\dfrac{3-1}{1.2.3}+\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}=\)

\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+\dfrac{1}{98.99}-\dfrac{1}{99.100}=\)

\(=\dfrac{1}{2}-\dfrac{1}{99.100}\)

c/

\(\dfrac{2}{5}.d=\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}+\dfrac{101-99}{99.100.101}=\)

\(=\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}+\dfrac{1}{99.100}-\dfrac{1}{100.101}=\)

\(=\dfrac{1}{2.3}-\dfrac{1}{100.101}\Rightarrow d=\left(\dfrac{1}{2.3}-\dfrac{1}{100.101}\right):\dfrac{2}{5}\)

Đặt A=1/1.2.3+1/2.3.4+...+1/99.100.101 

2A=2/1.2.3+2/2.3.4+...2/99.100.101

2A=3-1/1.2.3+4-2/2.3.4+...+101-99/99.100.101

2A=3/1.2.3-1/1.2.3+4/2.3.4-2/2.3.4+...+101/99.100.101-99/99.100.101

2A=1/1.2-1/2.3+1/2.3-1/3.4+...+1/99.100-1/100.101

2A=1/2-1/10100

mình ko biết 

  A = 1.2.3 + 2.3.4 + 3.4.5 + ... + 99.100.101

4A = 4.(1.2.3 + 2.3.4 + 3.4.5 + ... + 99.100.101)

     = 1.2.3.(4-0) + 2.3.4.(5-1) + 3.4.5.(6-2) + ... + 99.100.101.(102-98)

     = 1.2.3.4 - 1.2.3.4 + 2.3.4.5 - 2.3.4.5 + 3.4.5.6 - 3.4.5.6 + ... + 98.99.100.101 - 98.99.100.101 + 99.100.101.102

4A = 99.100.101.102

  A = 99.100.101.102 : 4 

  A = 25497450

Đặt \(A=1.2.3+2.3.4+3.4.5+...+99.100.101\)

\(\Rightarrow4A=1.2.3.4+2.3.4.4+...+99.100.101.4\)

\(=1.2.3\left(4-0\right)+2.3.4\left(5-1\right)+...+99.100.101\left(102-98\right)\)

\(=\left(1.2.3.4+2.3.4.5+...+99.100+101.102\right)-\left(0.1.2.3+1.2.3.4+...+98.99.100.101\right)\)

\(=99.100.101.102-0.1.2.3\)

\(=101989800\)

\(\Rightarrow A=101989800:4=25497450\)

Vậy \(A=25497450.\)