Giải thích các bước giải:

Ta có:

$B=1+\dfrac12+\dfrac13+...+\dfrac1{100}$

$\to B=(1+\dfrac1{100})+(\dfrac12+\dfrac1{99})+(\dfrac13+\dfrac1{98})+....+(\dfrac1{50}+\dfrac1{51})$

$\to B=(1+\dfrac1{100})+(\dfrac12+\dfrac1{99})+(\dfrac13+\dfrac1{98})+....+(\dfrac1{50}+\dfrac1{51})$

$\to B=\dfrac{101}{1\cdot 100}+\dfrac{101}{2\cdot 99}+\dfrac{101}{3\cdot 98}+...+\dfrac{101}{50\cdot 51}$

$\to B=101\cdot (\dfrac{1}{1\cdot 100}+\dfrac{1}{2\cdot 99}+\dfrac{1}{3\cdot 98}+...+\dfrac{1}{50\cdot 51})$

Lại có:

$A=1\cdot 2\cdot 3\cdot 4\cdots 99\cdot 100\cdot (1+\dfrac12+\dfrac13+...+\dfrac1{100})$

$\to A=1\cdot 2\cdot 3\cdot 4\cdots 99\cdot 100\cdot 101\cdot (\dfrac{1}{1\cdot 100}+\dfrac{1}{2\cdot 99}+\dfrac{1}{3\cdot 98}+...+\dfrac{1}{50\cdot 51})$

$\to A= 101\cdot (1\cdot 2\cdot 3\cdot 4\cdots 99\cdot 100\cdot (\dfrac{1}{1\cdot 100}+\dfrac{1}{2\cdot 99}+\dfrac{1}{3\cdot 98}+...+\dfrac{1}{50\cdot 51}))$

Mà $ (1\cdot 2\cdot 3\cdot 4\cdots 99\cdot 100\cdot (\dfrac{1}{1\cdot 100}+\dfrac{1}{2\cdot 99}+\dfrac{1}{3\cdot 98}+...+\dfrac{1}{50\cdot 51}))\in Z$

$\to A\quad\vdots\quad 101$

A=1.2.3.....99.100.(1+1/2+1/3+...+1/99+1/100)

  =1.2.3.....99.100.[(1+1/100)+(1/2+1/99)+(1/3+1/98)+...+(1/50+1/51)]

  =1.2.3.....99.100.(101/100+101/2.99+101/3.98+...+101/50.51)

  =1.2.3.....99.100.101.(1/100+1/2.99+...+1/50.51)

Vì 101:101⇒A:101(đpcm)

Vậy số tự nhiên A chia hết cho 101