Ta có:
$\dfrac{n}{3} + \dfrac{n^2}{2} + \dfrac{n^3}{6}$
$= \dfrac{2n}{6} + \dfrac{3n^2}{6} + \dfrac{n^3}{6}$
$= \dfrac{2n+3n^2+n^3}{6}$
$= \dfrac{2n+2n^2+n^2+n^3}{6}$
$= \dfrac{2n.(1+n) + n^2.(1 + n)}{6}$
$= \dfrac{(n+1).(2n+n^2)}{6}$
$= \dfrac{n.(n+1).(n+2)}{6}$
Vì : $n;n+1;n+2$ là $3$ số liên tiếp
$⇒$ Tích $3$ số này chia hết cho $6$
$⇒$ $\dfrac{n}{3} + \dfrac{n^2}{2} + \dfrac{n^3}{6}$ ($đ.p.c.m$).
