m×n= r×r= bx2.5= h×1= 省略乘号m×n= r×r= bx2.5= h×1= 省略乘号写出上面式.

m×n= r×r= bx2.5= h×1= 省略乘号m×n= r×r= bx2.5= h×1= 省略乘号写出上面式. 【急】三个组合恒等式求证明C(r,r)+C(r,r+1)+C(r,r+2)+,+C(r,n)=C(r+1,n+1)C(r,m)*C(0,n)+C(r-1.m)*C(1,n)+.+C(0.m)*C(r,n)=C(r,m+n)[C(0,n)]^2+[C(1,n)]^2+.=C(n,2n) r=m MOD n matlab 中总是提醒括号数不对,但我检查了好多遍都是对的啊.为什么c = [-n*H*(n+m)*[r_c*H^2+r*hu^2+r_s*hs^2-r*H*(alpha*hu+hd)]-m*H*(n+m)*[r_c*H^2+r*hd^2-r*H*(alpha*hu+hd)]-r_c*H^3*(n+m)*(m-n)+2*r*hu^3-r*hu^2*n*[3*H*(n+m)-2*hu*n]-2 #include main() { float r,h,s; scanf(%f %f,&r,&h); s=1/3*(3.14*r*r*h); printf(s=%f ,s) 定理：A是m*n矩阵,r(A)=r i=((1+r)n+m-1)/(n+m) r=5.4% n=2 m=1 求i 设A是m*n矩阵,且R(A)=r,则当r=m,r=n,m=n,r 1　m,l,f,i,a,y=?2 n,l,o,i=?3 j,i,e,u,c=?4 a,f,e,r,t,h=?5 m,o,n,l,e=?6 r,f,i,n,d,e=?7 t,r,e,i,s,s=?8 h,t,s,i=? 求M和N最大公约数?#include<stdio.h>void main(){ int m, n, r; scanf("&d,%d",&m,&n); do { m%n==r; if(r==0); printf("big is %d ",n); else { n==r; m==n; m%n==r; } }while(r==0); if(r==0); printf(& C(m,n+r+1)=C(m,n)C(0,r)+C(m-1,n-1)C(1,r+1)+...+C(0,n-m)C(m,r+m) 其组合意义证明 已知等比数列an,m=(sn-r,2^n-1),N=(2.1)且向量m//n求常数r 1/(1-r)=1+r+r^2+r^3+…+r^n其中|r|＜1怎么证? [m,n]=size(x); r=ones(m,n); 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)+2C(n,2)+…+nC(n,n)=n2^(n-1)还有：C(m,r)*C(n,0)+C(m,r-1)*C(n,1)+…+C(m,0)*C(n,r)=C(m+n,r) (C(n,o))^2+(C(n,1))^2+(C(n,2))^2+(C(n,3))^2+…+(C(n,n))^2=C(2n,n) 圆台体积计算?V=1/3*π*H*(R^2+r^2+R*r) (R为大圆半径,r为小圆半径,H为圆台高度） 设m*r矩阵F是列满秩,r*n矩阵G是行满秩,证明秩(FG)=r, 用matlab画如下式子的图像!应该怎样写程序?R=2＊r.＊sin(acos((r^2+(r+h).^2－R.^2./4)./(2*r.＊(r+h))))./cos(pi/180.*j+acos((r^2+(r+h).^2-R.^2./4)./(2*r.*(r+h)))); 其中r=6370,j=5,h=60:90 非常急,