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=2666666000
Có công thức như sau
1x2+2x3+3x4+...+nx(n+1)=nx(n+1)x(n+2):3
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\frac{A}{B}=\frac{\frac{2000}{1}+\frac{1999}{2}+\frac{1998}{3}+...+\frac{1}{2000}+2000}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}\)
\(\Leftrightarrow\frac{A}{B}=\frac{\left(\frac{2000}{1}+1\right)+\left(\frac{1999}{2}+1\right)+\left(\frac{1998}{3}+1\right)+...+\left(\frac{1}{2000}+1\right)+2000+1}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}\)
\(\Leftrightarrow\frac{A}{B}=\frac{\frac{2001}{1}+\frac{2001}{2}+\frac{2001}{3}+...+\frac{2001}{2000}+2001}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}\)
\(\Leftrightarrow\frac{A}{B}=\frac{2001\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}\right)}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}\)
\(\Leftrightarrow\frac{A}{B}=2001\)
bn cộng trên tử rồi thì phải trừ đi chứ ko phân số sẽ thay đổi
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt A=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2000}}{\frac{1999}{1}+\frac{1998}{2}+....+\frac{1}{1999}}\)
Xét mẫu số:
\(\frac{1999}{1}+\frac{1998}{2}+\frac{1997}{3}+\frac{1996}{4}+....+\frac{1}{1999}\)
=\(\left(\frac{1998}{2}+1\right)+\left(\frac{1997}{3}+1\right)+\left(\frac{1996}{4}+1\right)+....+\left(\frac{1}{1999}+1\right)+1\)
=\(\frac{2000}{2}+\frac{2000}{3}+\frac{2000}{4}+....+\frac{2000}{1999}+\frac{2000}{2000}\)
= 2000\(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{1999}+\frac{1}{2000}\right)\)
=> A = \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2000}}{2000\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2000}\right)}\)
=> A = \(\frac{1}{2000}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}{\frac{1999}{1}+\frac{1998}{2}+\frac{1997}{3}+....+\frac{1}{1999}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2000}}{1+\left(\frac{1998}{2}+1\right)+\left(\frac{1997}{3}+1\right)+....+\left(\frac{1}{1999}+1\right)}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}{\frac{2000}{2}+\frac{2000}{3}+\frac{2000}{4}+....+\frac{2000}{2000}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}{2000\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}\right)}\)
\(=\frac{1}{2000}\)