\(E=\frac{1}{1.2}-\frac{1}{1.2.3}+\frac{1}{2.3}-\frac{1}{2.3.4}+....+\frac{1}{99.100}-\frac{1}{99.100.101}\)

\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{99.100.101}\right)\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}=\frac{99}{100}\)

\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{99.100.101}\)

\(=\frac{1}{2}\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{101-99}{99.100.101}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)=\frac{5049}{20200}\)

Suy ra \(E=A-B=\frac{99}{100}-\frac{5049}{20200}=\frac{14949}{20200}\)

\(\frac{14949}{20200}\)

\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{99.100.101}\right)\)

\(=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)

\(=\left(1-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)\)

\(=\frac{99}{100}-\frac{1}{2}\cdot\frac{5049}{10100}=\frac{99}{100}-\frac{5049}{20200}=\frac{14949}{20200}\)

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

Nhân cả hai vế với 2

\(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}=\frac{2}{x}\left(\frac{1}{1.2}-\frac{1}{99.100}\right).\)

Xét vế trái

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

\(VT=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\)

\(VT=\frac{1}{1.2}-\frac{1}{99.100}\)

\(\Rightarrow\frac{2}{x}=1\Rightarrow x=2\)

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