组合数学中恒等式的证明:1、Σ(i=0,n)i^2*C(n,i)=n*(n+1)*2^(n-2);
来源:学生作业帮 编辑:搜狗做题网作业帮 分类:数学作业 时间:2024/07/19 03:03:17
组合数学中恒等式的证明:1、Σ(i=0,n)i^2*C(n,i)=n*(n+1)*2^(n-2);
还有一个:Σ(i=0,n)(1/(i+1)(i+2))C(n,i)=(2^(n+2)-n-3)/((n+1)(n+2))麻烦给出详解,
还有一个:Σ(i=0,n)(1/(i+1)(i+2))C(n,i)=(2^(n+2)-n-3)/((n+1)(n+2))麻烦给出详解,
![组合数学中恒等式的证明:1、Σ(i=0,n)i^2*C(n,i)=n*(n+1)*2^(n-2);](/uploads/image/z/7963579-19-9.jpg?t=%E7%BB%84%E5%90%88%E6%95%B0%E5%AD%A6%E4%B8%AD%E6%81%92%E7%AD%89%E5%BC%8F%E7%9A%84%E8%AF%81%E6%98%8E%EF%BC%9A1%E3%80%81%CE%A3%28i%3D0%2Cn%29i%5E2%2AC%28n%2Ci%29%3Dn%2A%28n%2B1%29%2A2%5E%28n-2%29%3B)
第一个,利用 (1+x)^n=Σ(i=0,n) C(n,i)*x^i,两边对x求导,得:
n*(1+x)^(n-1)=Σ(i=1,n) i*C(n,i)*x^(i-1).两边同乘以x,得:
n*x*(1+x)^(n-1)=Σ(i=1,n) i*C(n,i)*x^i.两边再对x求导,得:
n*(1+x)^(n-1)+n*x*(n-1)*(1+x)^(n-2)=Σ(i=1,n) i^2*C(n,i)*x^(i-1).令x=1,整理,得证.
第二个,利用 A:Σ(i=0,n) C(n,i)=2^n 和 B:Σ(i=1,n) i*C(n,i)=n*2^(n-1).
左式=Σ(i=0,n) (1/(i+1)-1/(i+2))*C(n,i)
=(1-1/2)*C(n,0)+(1/2-1/3)*C(n,1)+(1/3-1/4)*C(n,2)+...+(1/n-1/(n+1))*C(n,n-1)+(1/(n+1)-1/(n+2))*C(n,n)
=C(n,0)-(1/(n+2))*C(n,n)+(1/2)*(C(n,1)-C(n,0))+(1/3)*(C(n,2)-C(n,1))+...+(1/(n+1))*(C(n,n)-(n,n-1))
=1-1/(n+2)+Σ(i=1,n) (1/(i+1))*(C(n,i)-C(n,i-1))
=(n+1)/(n+2)+Σ (1/(i+1))*(n!/ (i!*(n-i)!) - n!/ ((i-1)!*(n-i+1)!))
=(n+1)/(n+2)+Σ (1/(i+1))*((n-i-1-i)*n!) / (i!*(n-i+1)!)
=(n+1)/(n+2)+Σ ((n-2*i+1)/(n+1)*(n+2))*((n+2)!/ ((i+1)!*(n-i+1)!))
=(n+1)/(n+2)+1/((n+1)(n+2)) Σ (n+3-2*i-2)*C(n+2,i+1)
=(n+1)/(n+2)+(n+3)/((n+1)*(n+2)) Σ C(n+2,i+1) - 2/((n+1)*(n+2)) Σ (i+1)*C(n+2,i+1)
利用A化简第二项
=(n+1)/(n+2)+((n+3)*(2^(n+2)-n-4)) / ((n+1)*(n+2)) - 2/((n+1)*(n+2)) Σ(i=2,n+1) i*C(n+2,i)
利用B化简第三项
=(n+1)/(n+2)+((n+3)*(2^(n+2)-n-4))/((n+1)*(n+2)) - (2*((n+2)*2^(n+1)-2*n-4)) / ((n+1)*(n+2))
然后化简一下,就得证啦.
n*(1+x)^(n-1)=Σ(i=1,n) i*C(n,i)*x^(i-1).两边同乘以x,得:
n*x*(1+x)^(n-1)=Σ(i=1,n) i*C(n,i)*x^i.两边再对x求导,得:
n*(1+x)^(n-1)+n*x*(n-1)*(1+x)^(n-2)=Σ(i=1,n) i^2*C(n,i)*x^(i-1).令x=1,整理,得证.
第二个,利用 A:Σ(i=0,n) C(n,i)=2^n 和 B:Σ(i=1,n) i*C(n,i)=n*2^(n-1).
左式=Σ(i=0,n) (1/(i+1)-1/(i+2))*C(n,i)
=(1-1/2)*C(n,0)+(1/2-1/3)*C(n,1)+(1/3-1/4)*C(n,2)+...+(1/n-1/(n+1))*C(n,n-1)+(1/(n+1)-1/(n+2))*C(n,n)
=C(n,0)-(1/(n+2))*C(n,n)+(1/2)*(C(n,1)-C(n,0))+(1/3)*(C(n,2)-C(n,1))+...+(1/(n+1))*(C(n,n)-(n,n-1))
=1-1/(n+2)+Σ(i=1,n) (1/(i+1))*(C(n,i)-C(n,i-1))
=(n+1)/(n+2)+Σ (1/(i+1))*(n!/ (i!*(n-i)!) - n!/ ((i-1)!*(n-i+1)!))
=(n+1)/(n+2)+Σ (1/(i+1))*((n-i-1-i)*n!) / (i!*(n-i+1)!)
=(n+1)/(n+2)+Σ ((n-2*i+1)/(n+1)*(n+2))*((n+2)!/ ((i+1)!*(n-i+1)!))
=(n+1)/(n+2)+1/((n+1)(n+2)) Σ (n+3-2*i-2)*C(n+2,i+1)
=(n+1)/(n+2)+(n+3)/((n+1)*(n+2)) Σ C(n+2,i+1) - 2/((n+1)*(n+2)) Σ (i+1)*C(n+2,i+1)
利用A化简第二项
=(n+1)/(n+2)+((n+3)*(2^(n+2)-n-4)) / ((n+1)*(n+2)) - 2/((n+1)*(n+2)) Σ(i=2,n+1) i*C(n+2,i)
利用B化简第三项
=(n+1)/(n+2)+((n+3)*(2^(n+2)-n-4))/((n+1)*(n+2)) - (2*((n+2)*2^(n+1)-2*n-4)) / ((n+1)*(n+2))
然后化简一下,就得证啦.
组合数学中恒等式的证明:1、Σ(i=0,n)i^2*C(n,i)=n*(n+1)*2^(n-2);
代数恒等式证明1/(1*n) +1/(2*(n-1))+1/(3*(n-2))+.+1/(i*(n-i+1))+.+1/
组合恒等式的证明:C(r,r)+C(r+1,r)+C(r+2,r)+…+C(n,r)=C(n+1,r+1) C(n,1)
求证c(0,n)+2c(i,n)+.+(n+1c(n,n)=(n+2)*2的n-1次方
【线性代数】证明恒等式,其中ai不等于0,i=1,2,.n
用数学归纳法证明恒等式:1+2+3+...+n^2 = (n^4+n^2)/2
组合:C(n,0)+C(n,1)+……+C(n,n)=n^2
组合猜想C(0,n)+C(1,n)+C(2,n)+C(3,n)+.+C(n,n) n∈N*的值,并证明你的结论
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(i,n)=2^n如何证明
组合数的公式,i*C(n,i)=n*C(n-1,i-1),这个公式该怎么理解?