组合恒等式2 五个基本的组合恒等式更复杂的技巧与例题

组合恒等式2 五个基本的组合恒等式复杂的例题

使用两个数列的递推关系证明恒等式
证明相反数与自身相等以说明恒等式为0
组合数倒数的裂项

这一讲继续讨论用五个基本的组合恒等式证明组合恒等式的方法。先复习一下基本恒等式：

等式一组合数的对称性
Cnk=Cnn−kC_n^k = C_n^{n-k}Cnk=Cnn−k

等式二杨辉三角
Cnk=Cn−1k+Cn−1k−1C_n^k = C_{n-1}^k + C_{n-1}^{k-1}Cnk=Cn−1k+Cn−1k−1

等式三多项式系数的唯一性
CnkCkm=CnmCn−mk−m=Cnk−mCn−k+mm,m≤k≤nC_n^kC_k^m = C_n^{m}C_{n-m}^{k-m} = C_n^{k-m}C_{n-k+m}^m,m\le k \le nCnkCkm=CnmCn−mk−m=Cnk−mCn−k+mm,m≤k≤n

特例，如果m=1m=1m=1，
kCnk=nCn−1k−1kC_n^k = nC_{n-1}^{k-1}kCnk=nCn−1k−1

等式四二项式定理
(x+y)n=∑i=0nCnixiyn−i(x+y)^n = \sum_{i=0}^n C_n^ix^iy^{n-i}(x+y)n=i=0∑nCnixiyn−i

特例，如果x=y=1x=y=1x=y=1，则
∑i=0nCni=2n\sum_{i=0}^n C_n^i = 2^ni=0∑nCni=2n

如果x=1,y=−1x=1,y=-1x=1,y=−1，则
∑i=0n(−1)iCni=0\sum_{i=0}^n (-1)^iC_n^i = 0i=0∑n(−1)iCni=0

使用两个数列的递推关系证明恒等式

上一讲介绍的例5和例6中，定义待证的和式为一个数列，用等式二裂项之和得到的两项都可以表示为这个数列的某种函数，从而得到这个数列的递推关系（一般是一阶递推）。但是当裂项得到的两项中有一项不能写成数列的函数时，我们可以再定义一个新的数列，然后对新的数列再用等式二裂项（一般可以得到第一个数列的二阶递推）。

例1 ∑k=0n(−1)k22n−2kC2n−k+1k=n+1\sum_{k=0}^n (-1)^k 2^{2n-2k}C_{2n-k+1}^k = n+1∑k=0n(−1)k22n−2kC2n−k+1k=n+1
证明
定义an=∑k=0n(−1)k22n−2kC2n−k+1ka_n = \sum_{k=0}^n (-1)^k 2^{2n-2k}C_{2n-k+1}^kan=∑k=0n(−1)k22n−2kC2n−k+1k，用等式二裂项，
an=22n+∑k=1n(−1)k22n−2k(C2n−kk+C2n−kk−1)=∑k=0n(−1)k22n−2kC2n−kk−∑j=0n−1(−1)j22n−2jC2(n−1)−j+1ja_n = 2^{2n}+ \sum_{k=1}^{n} (-1)^k 2^{2n-2k}(C_{2n-k}^k + C_{2n-k}^{k-1}) \\ = \sum_{k=0}^{n} (-1)^k 2^{2n-2k}C_{2n-k}^k - \sum_{j=0}^{n-1} (-1)^j 2^{2n-2j} C_{2(n-1)-j+1}^{j}an=22n+k=1∑n(−1)k22n−2k(C2n−kk+C2n−kk−1)=k=0∑n(−1)k22n−2kC2n−kk−j=0∑n−1(−1)j22n−2jC2(n−1)−j+1j

定义bn=∑k=0n(−1)k22n−2kC2n−kkb_{n} = \sum_{k=0}^{n} (-1)^k 2^{2n-2k}C_{2n-k}^kbn=∑k=0n(−1)k22n−2kC2n−kk，从而
bn=an+an−1b_{n} = a_{n} + a_{n-1}bn=an+an−1

用等式二对bnb_nbn裂项，
bn=22n+∑k=1n−1(−1)k22n−2k(C2n−k−1k+C2n−k−1k−1)+(−1)n=∑k=0n−1(−1)k22n−2kC2n−k−1k−22∑j=0n−1(−1)j22n−2jC2(n−1)−jj=4an−1−bn−1b_{n} = 2^{2n} + \sum_{k=1}^{n-1} (-1)^k 2^{2n-2k}(C_{2n-k-1}^k + C_{2n-k-1}^{k-1}) +(-1)^n\\ = \sum_{k=0}^{n-1}(-1)^k 2^{2n-2k}C_{2n-k-1}^k - 2^2 \sum_{j=0}^{n-1} (-1)^j 2^{2n-2j} C_{2(n-1)-j}^{j} \\ =4a_{n-1} - b_{n-1}bn=22n+k=1∑n−1(−1)k22n−2k(C2n−k−1k+C2n−k−1k−1)+(−1)n=k=0∑n−1(−1)k22n−2kC2n−k−1k−22j=0∑n−1(−1)j22n−2jC2(n−1)−jj=4an−1−bn−1

结合这两个递推式可得ana_nan的二阶线性递推式
an−an−1=an−1−an−2=1an=a0+(a1−a0)+⋯+(an−an−1)=1+na_{n} - a_{n-1} = a_{n-1} - a_{n-2} = 1 \\ a_n = a_0 + (a_1 - a_0) + \cdots + (a_n - a_{n-1}) = 1 + nan−an−1=an−1−an−2=1an=a0+(a1−a0)+⋯+(an−an−1)=1+n

证明相反数与自身相等以说明恒等式为0

有一些组合恒等式是恒等于0的，我们就可以尝试通过证明它与它的相反数相等来说明它的确恒等于0。
例2 ∑k=02n−1(−1)k+1(k+1)(C2nk)−1=0\sum_{k=0}^{2n-1}(-1)^{k+1}(k+1)(C_{2n}^k)^{-1}=0∑k=02n−1(−1)k+1(k+1)(C2nk)−1=0
证明
定义an=∑k=02n−1(−1)k+1(k+1)(C2nk)−1a_n=\sum_{k=0}^{2n-1}(-1)^{k+1}(k+1)(C_{2n}^k)^{-1}an=∑k=02n−1(−1)k+1(k+1)(C2nk)−1，考虑ana_nan的相反数，
−an=∑k=02n−1(−1)k(k+1)(C2nk)−1=∑k=02n−1(−1)2n−k(2n−k)(C2n2n−k−1)−1-a_n = \sum_{k=0}^{2n-1}(-1)^{k}(k+1)(C_{2n}^k)^{-1}=\sum_{k=0}^{2n-1}(-1)^{2n-k}(2n-k)(C_{2n}^{2n-k-1})^{-1}−an=k=0∑2n−1(−1)k(k+1)(C2nk)−1=k=0∑2n−1(−1)2n−k(2n−k)(C2n2n−k−1)−1

ana_nan与−an-a_n−an的区别仅仅只是对这2n2n2n项的求和顺序不一样，做指标变换j=2n−k−1j=2n-k-1j=2n−k−1就会发现an=−ana_n=-a_nan=−an，所以an=0a_n=0an=0。这里用到了两个有趣的对称性，首先是k+(2n−k)=2nk+(2n-k)=2nk+(2n−k)=2n偶数，则kkk与2n−k2n-k2n−k同为偶数或者同为奇数，所以(−1)k=(−1)2n−k(-1)^k=(-1)^{2n-k}(−1)k=(−1)2n−k，另外就是
(k+1)(C2nk)−1=(2n−k)(C2n2n−k−1)−1(k+1)(C_{2n}^k)^{-1}=(2n-k)(C_{2n}^{2n-k-1})^{-1}(k+1)(C2nk)−1=(2n−k)(C2n2n−k−1)−1

用一次等式三特例，再用一次等式一，然后再用一次等式三特例，
(k+1)(C2nk)−1=(2n+1)(C2n+1k+1)−1=(2n+1)(C2n+12n−k)−1=(2n−k)(C2n2n−k−1)−1(k+1)(C_{2n}^k)^{-1}=(2n+1)(C_{2n+1}^{k+1})^{-1} = (2n+1)(C_{2n+1}^{2n-k})^{-1}=(2n-k)(C_{2n}^{2n-k-1})^{-1}(k+1)(C2nk)−1=(2n+1)(C2n+1k+1)−1=(2n+1)(C2n+12n−k)−1=(2n−k)(C2n2n−k−1)−1

评注
1、这种证明方法的逻辑其实是
∑i=1nf(i)=∑i=1nf(n−i)=∑i=1n−f(i)\sum_{i=1}^nf(i) = \sum_{i=1}^nf(n-i) = \sum_{i=1}^n-f(i)i=1∑nf(i)=i=1∑nf(n−i)=i=1∑n−f(i)

一个可能的充分条件是下面这种关系
−f(i)=f(n−i)-f(i)=f(n-i)−f(i)=f(n−i)

几何解释就是(i,f(i))(i,f(i))(i,f(i))与(n−i,f(n−i))(n-i,f(n-i))(n−i,f(n−i))关于(n2,f(n2))(\frac{n}{2},f(\frac{n}{2}))(2n,f(2n))中心对称。

2、尝试对组合数用一次等式三特例，再用一次等式一，然后再用一次等式三特例，
Cnk=nkCn−1k−1=nkCn−1n−k=nkn−k+1nCnn−k+1⇒kCnk=(n−k+1)Cnn−k+1=[n−(k−1)]Cnk−1Cnk=k+1n+1Cn+1k+1=k+1n+1Cn+1n−k=k+1n+1n+1n−kCnn−k−1⇒(n−k)Cnk=(k+1)Cnn−k−1=(k+1)Cnk+1C_n^k = \frac{n}{k}C_{n-1}^{k-1}=\frac{n}{k}C_{n-1}^{n-k}=\frac{n}{k} \frac{n-k+1}{n}C_{n}^{n-k+1} \\ \Rightarrow kC_n^k = (n-k+1)C_{n}^{n-k+1} = [n-(k-1)]C_n^{k-1} \\ C_n^k = \frac{k+1}{n+1}C_{n+1}^{k+1}=\frac{k+1}{n+1}C_{n+1}^{n-k}=\frac{k+1}{n+1} \frac{n+1}{n-k}C_{n}^{n-k-1} \\ \Rightarrow (n-k)C_n^k = (k+1)C_{n}^{n-k-1} = (k+1)C_n^{k+1}Cnk=knCn−1k−1=knCn−1n−k=knnn−k+1Cnn−k+1⇒kCnk=(n−k+1)Cnn−k+1=[n−(k−1)]Cnk−1Cnk=n+1k+1Cn+1k+1=n+1k+1Cn+1n−k=n+1k+1n−kn+1Cnn−k−1⇒(n−k)Cnk=(k+1)Cnn−k−1=(k+1)Cnk+1

这些式子提供对组合数关于kkk的递推关系，等式三特例提供的是关于n,kn,kn,k同时增减的递推关系。

3、等式三的倒数依然成立，
(CnkCkm)−1=(CnmCn−mk−m)−1=(Cnk−mCn−k+mm)−1,m≤k≤n(C_n^kC_k^m)^{-1} = (C_n^{m}C_{n-m}^{k-m})^{-1} = (C_n^{k-m}C_{n-k+m}^m)^{-1},m\le k \le n(CnkCkm)−1=(CnmCn−mk−m)−1=(Cnk−mCn−k+mm)−1,m≤k≤n

假设m=1m=1m=1，则
(kCnk)−1=(nCn−1k−1)−1⇔nCnk=kCn−1k−1(kC_n^k)^{-1} = (nC_{n-1}^{k-1})^{-1} \Leftrightarrow \frac{n}{C_n^k} = \frac{k}{C_{n-1}^{k-1}}(kCnk)−1=(nCn−1k−1)−1⇔Cnkn=Cn−1k−1k

组合数倒数的裂项

既然等式三可以推广到组合数的倒数的形式，等式二也可以尝试推广。想象一下杨辉三角中所有的数字都变成它的倒数，那么肯定是相邻位置从下层到上层数字逐渐变小，要把(Cnk)−1(C_n^k)^{-1}(Cnk)−1拆成两项，肯定需要找下层与之相邻的两个元素，也就是(Cn+1k)−1(C_{n+1}^k)^{-1}(Cn+1k)−1和(Cn+1k+1)−1(C_{n+1}^{k+1})^{-1}(Cn+1k+1)−1，计算这两项的和，尝试凑出(Cnk)−1(C_n^k)^{-1}(Cnk)−1
(Cn+1k)−1+(Cn+1k+1)−1=k!(n+1−k)!+(k+1)!(n−k)!(n+1)!=k!(n−k)!n!(n+1−k)+(k+1)n+1=n+2n+1(Cnk)−1(C_{n+1}^k)^{-1}+(C_{n+1}^{k+1})^{-1} = \frac{k!(n+1-k)!+(k+1)!(n-k)!}{(n+1)!} \\ = \frac{k!(n-k)!}{n!}\frac{(n+1-k)+(k+1)}{n+1} = \frac{n+2}{n+1}(C_n^k)^{-1}(Cn+1k)−1+(Cn+1k+1)−1=(n+1)!k!(n+1−k)!+(k+1)!(n−k)!=n!k!(n−k)!n+1(n+1−k)+(k+1)=n+1n+2(Cnk)−1

所以，组合数的倒数的裂项方式为
1Cnk=n+1n+2(1Cn+1k+1Cn+1k+1)\frac{1}{C_n^k} = \frac{n+1}{n+2}\left( \frac{1}{C_{n+1}^k} + \frac{1}{C_{n+1}^{k+1}} \right)Cnk1=n+2n+1(Cn+1k1+Cn+1k+11)

例3 ∑k=12n−1(−1)k−1(C2nk)−1=1n+1\sum_{k=1}^{2n-1}(-1)^{k-1}(C_{2n}^k)^{-1} = \frac{1}{n+1}∑k=12n−1(−1)k−1(C2nk)−1=n+11
证明
定义an=∑k=12n−1(−1)k−1(C2nk)−1a_n=\sum_{k=1}^{2n-1}(-1)^{k-1}(C_{2n}^k)^{-1}an=∑k=12n−1(−1)k−1(C2nk)−1，用上面的倒数的裂项公式，
∑k=12n−1(−1)k−11C2nk=∑k=12n−1(−1)k−12n+12n+2(1C2n+1k+1C2n+1k+1)=2n+12n+2(1C2n+11+1C2n+12−1C2n+12−1C2n+13+1C2n+13⋯)=2n+12n+2(12n+1+12n+1)=1n+1\sum_{k=1}^{2n-1}(-1)^{k-1}\frac{1}{C_{2n}^k} = \sum_{k=1}^{2n-1}(-1)^{k-1}\frac{2n+1}{2n+2}\left( \frac{1}{C_{2n+1}^k} + \frac{1}{C_{2n+1}^{k+1}} \right) \\ = \frac{2n+1}{2n+2} \left( \frac{1}{C_{2n+1}^1} + \frac{1}{C_{2n+1}^{2}} -\frac{1}{C_{2n+1}^{2}} - \frac{1}{C_{2n+1}^{3}} + \frac{1}{C_{2n+1}^{3}} \cdots \right) \\ = \frac{2n+1}{2n+2} \left( \frac{1}{2n+1} + \frac{1}{2n+1} \right) = \frac{1}{n+1}k=1∑2n−1(−1)k−1C2nk1=k=1∑2n−1(−1)k−12n+22n+1(C2n+1k1+C2n+1k+11)=2n+22n+1(C2n+111+C2n+121−C2n+121−C2n+131+C2n+131⋯)=2n+22n+1(2n+11+2n+11)=n+11