定理

①：类欧几里得算法

求$$f(a,b,c,n)=\sum _{i=0}^n\lfloor \frac{a\times i+b}{c}\rfloor$$

分类讨论
①：$a\ge c||b\ge c$
$$\sum _{i=0}^n\lfloor \frac{a\times i+b}{c}\rfloor =\sum _{i=0}^n(\lfloor \frac{(a\%c)i+(b\%c)}{c}\rfloor +i\lfloor \frac{a}{c}\rfloor +\lfloor \frac{b}{c}\rfloor )$$
$$=f(a\%c,b\%c,c,n)+\frac{n\times (n+1)}{2}\lfloor \frac{a}{c}\rfloor +n\lfloor \frac{b}{c}\rfloor$$
②：$a<c\&\&b<c$

Ⅰ $a=0\Rightarrow f(a,b,c,n)=0$
Ⅱ $a\ne 0$
$$\sum _{i=0}^n\lfloor \frac{a\times i+b}{c}\rfloor =\sum _{i=0}^n\sum _{j=0}^{\lfloor \frac{a\times i+b}{c}\rfloor -1}1$$$$=\sum _{j=0}^{\lfloor \frac{a\ast n+b}{c}\rfloor -1}\sum _{i=0}^n[j<\lfloor \frac{a\times i+b}{c}\rfloor ]$$$$=\sum _{j=0}^{\lfloor \frac{a\ast ni+b}{c}\rfloor -1}\sum _{i=0}^n[j<\lceil \frac{ai+b-c+1}{c}\rceil ]$$$$=\sum _{j=0}^{\lfloor \frac{a\times i+b}{c}\rfloor -1}\sum _{i=0}^n[cj<ai+b-c+1]$$$$=\sum _{j=0}^{\lfloor \frac{a\times i+b}{c}\rfloor -1}\sum _{i=0}^n[cj-b+c-1<ai]$$$$=\sum _{j=0}^{\lfloor \frac{a\ast n+b}{c}\rfloor -1}\sum _{i=0}^n[\lfloor \frac{cj-b+c-1}{a}\rfloor <i]$$$$=\sum _{j=0}^{\lfloor \frac{a\ast n+b}{c}\rfloor -1}n-\lfloor \frac{cj-b+c-1}{a}\rfloor$$$$=n\lfloor \frac{a\ast i+b}{c}\rfloor -f(c,-b+c-1,a,\lfloor \frac{a\ast n+b}{c}\rfloor -1)$$
又套用①情况求解

公式

①：等比数列求和

$$\sum _{k=1}^n{x}^k=\frac{x-x^{n+1}}{1-x}$$

令$S_n={\sum }_{k=1}^n{x}^k$
$$x\times S_n=x\sum _{k=1}^n{x}^k=\sum _{k=2}^{n+1}{x}^k$$
两式相减即可
$$S_n-x\times S_n=x^1-x^{n+1}$$$$S_n=\frac{x-x^{n+1}}{1-x}$$

②：等差数列一次方和

首项为$a$，公差为$d$，求前$n$项等差数的和
$$\sum _{i=1}^n(a+(i-1)d)$$

$$\underset{n}{\underbrace{a+(a+d)+(a+2d)+...+(a+(n-1)d)}}=a\cdot n+\underset{n-1}{\underbrace{d+2d+...(n-1)d}}$$
$$=a\cdot n+(1+2+...+n-1)d=a\cdot n+\frac{n(n-1)}{2}d$$

③：等差数列二次方和

首项为$a$，公差为$d$，求前$n$项等差数的平方的和
$$\sum _{i=1}^n(a+(i-1)d{)}^2$$

$$\underset{n}{\underbrace{a^2+(a+d{)}^2+(a+2d{)}^2+...+(a+(n-1)d{)}^2}}$$
$$=\underset{n}{\underbrace{a^2+(a^2+2ad+1^2{d}^2)+(a^2+4ad+2^2{d}^2)+...+(a^2+2(n-1)ad+(n-1{)}^2{d}^2)}}$$
$$=a^{2}n+\underset{n-1}{\underbrace{(2ad+4ad+...+2(n-1)ad)}}+\underset{n-1}{\underbrace{(1^2+2^2+...+(n-1{)}^2)d^2}}$$
$$=a^{2}n+(1+2+...+(n-1))2ad+\frac{(n-1)n(2n-1)}{6}{d}^2$$
$$=a^{2}n+n(n-1)ad+\frac{(n-1)n(2n-1)}{6}{d}^2$$

结论

①：$n\&1=1\Rightarrow 3|(2^n-2)$

如果$n$为奇数，有$2^n-2$则为$3$的倍数

将减法转化为加法，如果$2^n+1=3k$成立，那么结论成立，现在来证明“如果”
分享两种方法

法1
设$n=2m+1$
$$2^n+1=2^{2m+1}+1=(2^{2m+1}-2^{2m-1})+(2^{2m-1}-2^{2m-3})+...+(2^3-2^1)+(2-1)$$
每个括号里面的数都是$3$的倍数，得证

• 简单举例第一个括号
  $$2^{2m+1}-2^{2m-1}=2^{2m-1}\ast (2^2-1)=2^{2m-1}\ast 3$$

法2：数学归纳法
当$n=1$时，有$2^n+1=3$，满足条件
当$n=k$时，$k$为奇，假设有$2^k+1=3t$
当$n=k+2$时，则有$2^n+1=2^{k+2}+1=4\ast 2^k+1=4\ast (3t-1)+1=12t-3=3\ast (4t-1)$
证毕

②：$1^2+2^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$

有$(n+1{)}^3=n^3+3n^2+3n+1$
$$\{\begin{array}{l}{2}^3=1^3+3\ast 1^2+3\ast 1+1\\ 3^3=2^3+3\ast 2^2+3\ast 2+1\\ ...\\ n^3=(n-1{)}^3+3\ast (n-1{)}^2+3(n-1)+1\\ (n+1{)}^3=n^3+3\ast n^2+3\ast n+1\end{array}$$
等式左右两边$n$项全加起来
$$(n+1{)}^3=1^3+3\ast (1^2+2^2+...+n^2)+3\ast (1+2+...+n)+\underset{n}{\underbrace{1+1+...+1}}$$
$$\iff (1^2+2^2+...+n^2)=\frac{(n+1{)}^3-1^3-3\ast (1+2+...+n)-\underset{n}{\underbrace{1+1+...+1}}}{3}$$
$$\iff (1^2+2^2+...+n^2)=\frac{n^3+3\ast n^2+3\ast n+1-1-3\ast \frac{n(n+1)}{2}-n}{3}$$
$$\iff (1^2+2^2+...+n^2)=\frac{2n^3+6n^2+6n+2-2-3n^2-3n-2n}{6}$$
$$\iff (1^2+2^2+...+n^2)=\frac{2n^3+3n^2+n}{6}=\frac{n(2n^2+3n+1)}{6}=\frac{n(n+1)(2n+1)}{6}$$