Giải thích các bước giải:
Đặt $A_n=n(n+1)(n+2)(n+3)(n+4)(n+5)$
$+)n=1\rightarrow A_1=120\quad\vdots\quad 120$
Giả sử $n=k\rightarrow A_k\quad\vdots\quad 120$
Ta cần chứng minh $A_{k+1}\quad\vdots\quad 120$
$\rightarrow A_{k+1}=(k+1)(k+2)(k+3)(k+4)(k+5)(k+6)=k(k+1)(k+2)(k+3)(k+4)(k+5)+6(k+1)(k+2)(k+3)(k+4)(k+5)$
$\rightarrow A_{k+1}=A_k+6(k+1)(k+2)(k+3)(k+4)(k+5)$
Vì $k+1,k+2,k+3,k+4,k+5$ là 5 số tự nhiên liên tiếp
$\rightarrow (k+1)(k+2)(k+3)(k+4)(k+5)\quad\vdots\quad 4.5$
$\rightarrow (k+1)(k+2)(k+3)(k+4)(k+5)\quad\vdots\quad 20$
$\rightarrow 6(k+1)(k+2)(k+3)(k+4)(k+5)\quad\vdots\quad 120$
$\rightarrow A_{k+1}=A_k+6(k+1)(k+2)(k+3)(k+4)(k+5)\quad\vdots\quad 120$
$\rightarrow A_n\quad\vdots\quad 120\rightarrow đpcm$
