小知识点随记

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$f_n -n\times (n+1) /2 = (n+1) \times n \times (n-1) /3$