范德蒙德卷积,一个绝妙的证明
作者:互联网
范德蒙德卷积:
\[\sum_{i=0}^k\dbinom ni\dbinom m{k-i}=\dbinom{n+m}k \]怎么证呢?
常见证法:
- 组合意义(天地灭)
- OGF(天地灭灭灭灭灭)
一个绝妙的证明:
显而易见 \(\forall a,b\) .
\[(a+b)^n(a+b)^m=(a+b)^{n+m} \]用二项式定理展开:
\[\left[\sum_{k=0}^n\dbinom nka^kb^{n-k}\right]\left[\sum_{k=0}^m\dbinom nka^kb^{m-k}\right]=\sum_{k=0}^{n+m}\dbinom {n+m}ka^kb^{n+m-k} \]\[\sum_{i=0}^n\sum_{j=0}^m\dbinom ni\dbinom mja^ib^jb^{n-i}b^{m-j}=\sum_{k=0}^{n+m}\dbinom {n+m}ka^kb^{n+m-k} \]\[\sum_{i=0}^n\sum_{j=0}^m\dbinom ni\dbinom mja^{i+j}b^{n+m-i-j}=\sum_{k=0}^{n+m}\dbinom {n+m}ka^kb^{n+m-k} \]对于 LHS,令 \(k=i+j\),转而枚举 \(k\) .
\[\mathrm{LHS}=\sum_{k=0}^{n+m}\sum_{j=0}^m\dbinom n{k-j}\dbinom mja^{k}b^{n+m-k} \]对比系数(此处正确性在于 \(\forall a,b\) 原式均成立)
\[\sum_{j=0}^m\dbinom n{k-j}\dbinom mj=\dbinom {n+m}k \]证毕了!!!!!!!
标签:kb,ka,dbinom,ni,卷积,sum,绝妙,德蒙,mja 来源: https://www.cnblogs.com/CDOI-24374/p/16102266.html