Đáp án:
$A= \dfrac{1}{1008}+\dfrac{1}{1009}+...+\dfrac{1}{2014}$
Giải thích các bước giải:
Ta có :
$A = \dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{2013.2014} $
$ = 1- \dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{2013}-\dfrac{1}{2014}$
$ = (1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{2013})-(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{2014})$
$ = (1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{2014})-2.(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{2014})$
$ = (1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{2014}) - (1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1007})$
$= \dfrac{1}{1008}+\dfrac{1}{1009}+...+\dfrac{1}{2014}$