Giải thích các bước giải:
Ta có: S = $\frac{1}{1000.1998}$ + $\frac{1}{1001.1997}$ + ... + $\frac{1}{1998.1000}$
= $\frac{1}{2998}$.($\frac{1000+1998}{1000.1998}$ + $\frac{1001+1997}{1001.1997}$ + ... + $\frac{1998+1000}{1998.1000}$)
= $\frac{1}{2998}$.($\frac{1}{1000}$ + $\frac{1}{1998}$ + $\frac{1}{1001}$+ $\frac{1}{1997}$ ... + $\frac{1}{1998}$ + $\frac{1}{1000}$)
= $\frac{2}{2998}$.($\frac{1}{1000}$ + $\frac{1}{1001}$+ $\frac{1}{1002}$ ... + $\frac{1}{1998}$)
= $\frac{1}{1499}$.[(1+ $\frac{1}{2}$ + $\frac{1}{3}$ + ... + $\frac{1}{1998}$) - (1 + $\frac{1}{2}$ + $\frac{1}{3}$ + ... + $\frac{1}{999}$)]
= $\frac{1}{1499}$.[(1+ $\frac{1}{2}$ + $\frac{1}{3}$ + ... + $\frac{1}{1998}$) - 2.($\frac{1}{2}$ + $\frac{1}{4}$ + ... + $\frac{1}{1998}$)]
= $\frac{1}{1499}$.(1 - $\frac{1}{2}$ + $\frac{1}{3}$ - $\frac{1}{4}$ + ... + $\frac{1}{1997}$ - $\frac{1}{1998}$)
= $\frac{1}{1499}$.($\frac{1}{1.2}$ + $\frac{1}{3.4}$ + $\frac{1}{5.6}$ + ... + $\frac{1}{1997.1998}$)
