$\dfrac{1}{100}.99 - \dfrac{1}{99}.98 - \dfrac{1}{98}.97 - ... - \dfrac{1}{3}.2 - \dfrac{1}{2}.1$
$= \dfrac{1}{100}.99 - \dfrac{1}{99}.98 - \dfrac{1}{98}.97 - ... - \dfrac{1}{3}.2 - \dfrac{1}{2}.1$
$= \dfrac{99}{100}- \dfrac{98}{99} - \dfrac{97}{98} - ... - \dfrac{2}{3}- \dfrac{1}{2}$
$= (1- \dfrac{1}{100})- (1-\dfrac{1}{99}) - (1- \dfrac{1}{98}) - ... - (1-\dfrac{1}{3}) - (1- \dfrac{1}{2})$
$= 1- \dfrac{1}{100}- 1+ \dfrac{1}{99} - 1 + \dfrac{1}{98} - ... - 1+ \dfrac{1}{3} - 1+ \dfrac{1}{2}$
$= (1-1-1-1..-1)+ (- \dfrac{1}{100}+ \dfrac{1}{99} + \dfrac{1}{98} +...+ \dfrac{1}{3} + \dfrac{1}{2})$
$= -97 - \dfrac{1}{100}+ \dfrac{1}{99} + \dfrac{1}{98} +...+ \dfrac{1}{3} + \dfrac{1}{2}$
Có lẽ chỉ tính đc đến đây thôi nhỉ :-s