Ta có:

\(\frac{A}{B}=\frac{\frac{2000}{1}+\frac{1999}{2}+\frac{1998}{3}+...+\frac{1}{2000}+2000}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}\)

\(\Leftrightarrow\frac{A}{B}=\frac{\left(\frac{2000}{1}+1\right)+\left(\frac{1999}{2}+1\right)+\left(\frac{1998}{3}+1\right)+...+\left(\frac{1}{2000}+1\right)+2000+1}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}\)

\(\Leftrightarrow\frac{A}{B}=\frac{\frac{2001}{1}+\frac{2001}{2}+\frac{2001}{3}+...+\frac{2001}{2000}+2001}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}\)

\(\Leftrightarrow\frac{A}{B}=\frac{2001\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}\right)}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}\)

\(\Leftrightarrow\frac{A}{B}=2001\)

=2666666000

Có công thức như sau

1x2+2x3+3x4+...+nx(n+1)=nx(n+1)x(n+2):3

phần sau thì sao

Ta có Đặt B = \(\frac{1999}{1}+\frac{1998}{2}+...+\frac{1}{1999}\)(1999 số hạng)                                 

\(=\left(1+1+1+...+1\right)+\frac{1998}{2}+\frac{1997}{3}+...+\frac{1}{1999}\)(1999 số hạng 1)            

\(=1+\left(\frac{1998}{2}+1\right)+\left(\frac{1997}{3}+1\right)+...+\left(\frac{1}{1999}+1\right)\)(1998 cặp số)

 = \(\frac{2000}{2}+\frac{2000}{3}+...+\frac{2000}{1999}+\frac{2000}{2000}\)

= \(2000\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1999}+\frac{1}{2000}\right)\)

Khi đó \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}{\frac{1999}{1}+\frac{1998}{2}+...+\frac{1}{1999}}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}{2000\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}\right)}=\frac{1}{2000}\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-1998}{1999}\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{1998}{1999}=\dfrac{1}{1999}\)

\(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{1999}-1\right)\)

\(=\left(\dfrac{1}{2}-\dfrac{2}{2}\right)\cdot\left(\dfrac{1}{3}-\dfrac{3}{3}\right)\cdot\left(\dfrac{1}{4}-\dfrac{4}{4}\right)...\left(\dfrac{1}{1999}-\dfrac{1999}{1999}\right)\)

\(=\dfrac{1-2}{2}\cdot\dfrac{1-3}{3}\cdot\dfrac{1-4}{4}\cdot...\cdot\dfrac{1-1999}{1999}\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot\dfrac{-3}{4}\cdot...\cdot\dfrac{-1998}{1999}\)

\(=\dfrac{-1\cdot-2\cdot-3\cdot...\cdot-1998}{2\cdot3\cdot4\cdot...\cdot1999}\)

\(=\dfrac{1}{1999}\)

TẦM NHƯ HƠI CĂNG

\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}{\frac{1999}{1}+\frac{1998}{2}+\frac{1997}{3}+....+\frac{1}{1999}}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2000}}{1+\left(\frac{1998}{2}+1\right)+\left(\frac{1997}{3}+1\right)+....+\left(\frac{1}{1999}+1\right)}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}{\frac{2000}{2}+\frac{2000}{3}+\frac{2000}{4}+....+\frac{2000}{2000}}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}{2000\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}\right)}\)

\(=\frac{1}{2000}\)