Giải thích các bước giải:
Ta có :
$\dfrac{1}{1.2}+\dfrac{1}{3.4}+..+\dfrac{1}{199.200}$
$=\dfrac{2-1}{1.2}+\dfrac{4-3}{3.4}+..+\dfrac{200-199}{199.200}$
$=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...\dfrac{1}{199}-\dfrac{1}{200}$
$=(\dfrac{1}{1}+\dfrac{1}{3}+\dfrac{1}{5}+..+\dfrac{1}{199})-(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+..+\dfrac{1}{200})$
$=(\dfrac{1}{1}+\dfrac{1}{3}+\dfrac{1}{5}+..+\dfrac{1}{199}+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+..+\dfrac{1}{200})-2(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+..+\dfrac{1}{200})$
$=(\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+..+\dfrac{1}{199}+\dfrac{1}{200})-2(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+..+\dfrac{1}{200})$
$=(\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+..+\dfrac{1}{199}+\dfrac{1}{200})-(\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+..+\dfrac{1}{100})$
$=\dfrac{1}{101}+\dfrac{1}{102}+..+\dfrac{1}{200}$
$\to N=1$
