Giải thích các bước giải:
Ta có:
$A=\dfrac{1.2-1}{2!}+\dfrac{2.3-1}{3!}+\dfrac{3.4-1}{4!}+...+\dfrac{99.100-1}{100!}$
$\to A=\dfrac{1.2}{2!}-\dfrac{1}{2!}+\dfrac{2.3}{3!}-\dfrac{1}{3!}+\dfrac{3.4}{4!}-\dfrac{1}{4!}+....+\dfrac{99.100}{100!}-\dfrac{1}{100!}$
$\to A=(\dfrac{1.2}{2!}+\dfrac{2.3}{3!}+\dfrac{3.4}{4!}+....+\dfrac{99.100}{100!})-(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+....+\dfrac{1}{100!})$
$\to A=(1+1+\dfrac1{2!}+...+\dfrac{1}{98!})-(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+....+\dfrac{1}{100!})$
$\to A=1+1-\dfrac{1}{99!}-\dfrac{1}{100!}$
$\to A=2-\dfrac{1}{99!}-\dfrac{1}{100!}$
$\to A<2$
