Đáp án:
$\dfrac{99}{100}$
Giải thích các bước giải:
d)
$\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+ . . .+\dfrac{1}{9900}$
$=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+. . .+\dfrac{1}{99.100}$
$=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+ . . . +\dfrac{1}{99}-\dfrac{1}{100}$
$=1-\dfrac{1}{100}$
$=\dfrac{99}{100}$
