数学分析解题思路

吉米多维奇数学分析中一道习题的分享
证明:$1^2+2^2+\cdots +n^2=\frac{n(n+1)(2n+1)}{6}$
两种方法：

1. 数学归纳法
   证明: 当$n=1$时,等式显然成立
   设$n=k$时,等式成立,即$1^2+2^2+\cdots +k^2=\frac{k(k+1)(2k+1)}{6}+(k+1{)}^2$,则对于$n=k+1$时,有
   $$\begin{array}{rl}{1}^2+2^2+\cdots +k^2+(k+1{)}^2& =\frac{k(k+1)(2k+1)}{6}+(k+1{)}^2\\ & =\frac{1}{6}(k+1)[k(2k+1)+6(k+1)]=\frac{(k+1)[(k+1)+1][2(k+1)+1]}{6}\end{array}$$
   即对于$n=k+1$时,等式也成立,于是,对于任何正整数$n$,有$1^2+2^2+\cdots +n^2=\frac{n(n+1)(2n+1)}{6}$
   2.直接验证法
   首先$(n+1{)}^3=n^3+3n^2+3n+1$即$n^2=\frac{1}{3}[(n+1{)}^3-n^3-3n-1]$
   那么求和可以写为:
   $$\frac{1}{3}\left[2^3-1^3-3-1+3^3-2^3-3\ast 2-1+\cdots +(n+1{)}^3-n^3-3n-1\right]$$
   显然三次方项都被相互抵消，$-3-3\ast 2+3\ast 3-\cdots$是等差数列，$-1-1-1-1-\cdots$是常数列.那么上式改写为:
   $$\begin{array}{rl}& =\frac{1}{3}\left[(n+1{)}^3+1^3-n-\frac{3n(n+1)}{2}\right]\\ & =\frac{1}{3}\frac{3n^2+3n-2n-2}{2}（通分）\\ & =\frac{2n^3+3n^2+n}{6}\\ & =\frac{n(n+1)(2n+1)}{6}\end{array}$$
   上式得证