1. $\sum\limits_{i=1}^{n} i^2 = (2 \times n^3+3 \times n^2+n)/6$
证明一：
设$ f_n = \sum\limits_{i=1}^{n} i^2 $,$ S_n = \sum\limits_{i=1}^{n} i^3 $
$S_{n+1} = S_{n}+(i+1)^3 $
$S_{n+1} = \sum\limits_{i=0}^{n} (i+1)^3 $
$S_{n+1} = \sum\limits_{i=0}^{n} i^3 + 3 \times i^2 + 3 \times i +1$
$(n+1)^3 = \sum\limits_{i=0}^{n} 3 \times i^2 + 3 \times i +1$
$n^3 +3 \times n^2 + 3 \times n + 1 = 3 f_n + 3 \times n \times (n+1)/2 +n+1$
整理一下就行了。
证明二：
利用$\sum\limits_{i=1}^n C_i^k = C_{n+1}^{k+1}$(证明: $\sum\limits_{i=1}^n C_i^k = C_1^{k+1} + \sum\limits_{i=1}^n C_i^k = C_2^{k+1}+ \sum\limits_{i=2}^n C_i^k = C_{n+1}^{k+1}$)
那么$\sum\limits_{i=1}^n C_i^2 = C_{n+1}^3$,
然后得$\sum\limits_{i=1}^n i \times (i-1) /2 = (n+1) \times n \times(n-1) /6$
$\sum\limits_{i=1}^n i \times i - \sum\limits_{i=1}^n i = \sum\limits_{i=1}^n i \times i - n \times (n+1) /2$
也就是$f_n -n\times (n+1) /2 = (n+1) \times n \times (n-1) /3$
原文：https://www.cnblogs.com/hzoi-kx/p/11721861.html