C(n,0)+2C(n,1)+3C(n,2)+...+(n+1)C(n,n)=256,那么n=
热心网友
S=[C(n,0)+C(n,1)+C(n,2)+.....+C(n,n)]+[0C(n,0)+1C(n,1)+...+nC(n,n)]因为C(n,0)+C(n,1)+...+C(n,n)=2^n0C(n,0)+1C(n,1)+...+nC(n,n)=n×2^(n-1)所以S=2^n+n×2^(n-1)=256,即[2^(n-1)]×(2+n)=256易得:n=6
C(n,0)+2C(n,1)+3C(n,2)+...+(n+1)C(n,n)=256,那么n=
S=[C(n,0)+C(n,1)+C(n,2)+.....+C(n,n)]+[0C(n,0)+1C(n,1)+...+nC(n,n)]因为C(n,0)+C(n,1)+...+C(n,n)=2^n0C(n,0)+1C(n,1)+...+nC(n,n)=n×2^(n-1)所以S=2^n+n×2^(n-1)=256,即[2^(n-1)]×(2+n)=256易得:n=6