证明C(n+1,k)=C(n,k-1)+C(n,k) 及 C(n,r)*C(r,k)=C(n,k)*C(n-k,r-k)证明C(n+1,k)=C(n,k-1)+C(n,k)证明C(n,r)*C(r,k)=C(n,k)*C(n-k,r-k)

来源:学生作业帮助网 编辑:作业帮 时间:2024/04/28 02:16:58
证明C(n+1,k)=C(n,k-1)+C(n,k) 及 C(n,r)*C(r,k)=C(n,k)*C(n-k,r-k)证明C(n+1,k)=C(n,k-1)+C(n,k)证明C(n,r)*C(r,k)=C(n,k)*C(n-k,r-k)

证明C(n+1,k)=C(n,k-1)+C(n,k) 及 C(n,r)*C(r,k)=C(n,k)*C(n-k,r-k)证明C(n+1,k)=C(n,k-1)+C(n,k)证明C(n,r)*C(r,k)=C(n,k)*C(n-k,r-k)
证明C(n+1,k)=C(n,k-1)+C(n,k) 及 C(n,r)*C(r,k)=C(n,k)*C(n-k,r-k)
证明C(n+1,k)=C(n,k-1)+C(n,k)
证明C(n,r)*C(r,k)=C(n,k)*C(n-k,r-k)

证明C(n+1,k)=C(n,k-1)+C(n,k) 及 C(n,r)*C(r,k)=C(n,k)*C(n-k,r-k)证明C(n+1,k)=C(n,k-1)+C(n,k)证明C(n,r)*C(r,k)=C(n,k)*C(n-k,r-k)
(1)
C(n,k-1)+C(n,k) = n!/((k-1)!*(n-k+1)!) + n!/(k!*(n-k)!)
= n!*k/(k!*(n-k+1)!) +n!*(n-k+1)/(k!*(n-k+1)!)
= n!*(n+1)/(k!*(n-k+1)!)
= (n+1)!/(k!*(n-k+1)!)
= C(n+1,k)
(2)
C(n,r)*C(r,k) = n!/(r!*(n-r)!) * r!/(k!*(r-k)!) = n!/((n-r)!*(r-k)!*k!)
C(n,k)*C(n-k,r-k) = n!/(k!*(n-k)!) * (n-k)!/((r-k)!*(n-r)!) = n!/(k!*(r-k)!*(n-r)!)
∴ C(n,r)*C(r,k) = C(n,k)*C(n-k,r-k)

证明C(n+1,k)=C(n,k-1)+C(n,k) 及 C(n,r)*C(r,k)=C(n,k)*C(n-k,r-k)证明C(n+1,k)=C(n,k-1)+C(n,k)证明C(n,r)*C(r,k)=C(n,k)*C(n-k,r-k) 证明组合C(n-1,k)+C(n-2,k)+…+C(k+1,k)+C(k,k)=C(n,k+1) 证明n*(x+1)^(n-1)=Σ(k=0到n)k*c(n,k)*x^(k-1) 证明C(0,n)+C(1,n+1)+C(2,n+2)+...+C(k,n+k)=C(k,n+k+1) 不展开 用排列组合意义证明 C(n-1,k-1)C(n,k+1)C(n+1,k)=C(n-1,k)C(n,k-1)C(n+1,k+1) 怎么证明∑c(k,n)p^k*q^(n-k)=1= =对不起啊,题目问错了...应该是证明介个...∑[c(k,M)*c(n-k,N-M)]/c(n,N)=1 用加法原理证明,C(n+1,k)=C(n,k)+C(n,k-1)注:括号内左边为下标 ,右边为上标, 证明:(n+1)!/k!-n!/(k-1)!=(n-k+1)*n!/k!(k≤n) 试证明:∑(i=1到n)C(n,i)*k^(n-i)*k*i=n*k*(k+1)^(n-1)RT,求证明过程,要求看得明白 C(n,k)+C(n,k-1)C(m,1)+C(n,k-2)C(m,2)+.+C(m,k)=?求证:C(n,k)+C(n,k-1)C(m,1)+C(n,k-2)C(m,2)+.+C(m,k)=C(m+n,k) C语言for(n=k;1 试证明 x/[n(n+k)]=(x/k)[1/n-1/(n+k)] 证明组合恒等式:sum(k,0,m,C(n-k,m-k))=C(n+1,m) 至少2中方法! 求证两个组合恒等式(1)C(n,0)+C(n+1,1)+...+C(n+k,k)=C(n+k+1,k)(2)C(m,0)*C(n,k)+C(m,1)*C(n,k-1)+...+C(m,k)*C(n,0)=C(m+n,k) 证明 若x服从二项分布 则E(x)=npEX=∑kb(k;n,p)=∑k*C(k,n)p^kq^(n-k)=np∑C(k-1,n-1)p^(k-1)q^(n-1-k+1)=np∑C(k,n-1)p^kq^(n-1-k)=np∑b(k;n-1,p) ①=np ②前面的我都明白,请问怎 求证:Ck^K+Ck^(k+1)+Ck^(k+2)+Ck^(k+3)+...+Ck^(k+n)=C(k+1)^(k+n+1)(组合问题)急! 求∑C(k,n)*C(m-k,n),k=0,1,2.C表示数学中的组合 用数学归纳法证明(n+1)(n+2)······(n+n)=2^n·1·3·····(2n-1),从k到k+1,左边需要增乘的代A.2k+1 B.2(2k+1) C.(2k+1)/(k+1) D.(2k+3)/(k+1)