Đặt $f(k)=k(k+1)(k+2)$ suy ra $\Delta f= (k+1)(k+2)(k+3)-k(k+1)(k+2)=3(k+1)(k+2)$
Đặt $g(k)=\dfrac{k}{3}$ suy ra $\Delta g = \dfrac{1}{3}$
Khi đó:
$S=\sum\limits_{k=1}^{2010}k(k+1)(k+2)=\sum\limits_{k=1}^{2010} g(k)\Delta f=g(k)f(k)\left|\begin{matrix}{}^{2011} \\ {}_{k=1}\end{matrix}\right.-\sum\limits_{k=1}^{2010} f(k+1)\Delta g\quad\text{(SAI PHÂN TỪNG PHẦN)}$
$S=\dfrac{k^2(k+1)(k+2)}{3}\left|\begin{matrix}{}^{2011} \\ {}_{k=1}\end{matrix}\right.-\sum\limits_{k=1}^{2010} \dfrac{(k+1)(k+2)(k+3)}{3}$
$S=\dfrac{2011^2.2012.2013}{3}-\dfrac{1^2.2.3}{3}-\sum\limits_{k=2}^{2011} \dfrac{k(k+1)(k+2)}{3}$
$S=\dfrac{2011^2.2012.2013}{3}-\dfrac{1^2.2.3}{3}-\dfrac{2011.2012.2013}{3}+\dfrac{1.2.3}{3}-\sum\limits_{k=1}^{2010} \dfrac{k(k+1)(k+2)}{3}$
$S=\dfrac{2010.2011.2012.2013}{3}-\dfrac{1}{3}S$
$\dfrac{4S}{3}=\dfrac{2010.2011.2012.2013}{3}$
$S=\dfrac{2010.2011.2012.2013}{4}$
- C a c t u s yêu thích