化简:Cn0+1/2Cn1+1/3Cn2+...+1/(n+1)Cnn

1/(k+1)C(n,k)
=n!/(n-k)!k! * 1/(k+1)
=n!/(n-k)!(k+1)!
=(n+1)!/(n+1-k-1)!(k+1)! *1/(n+1)
=C(n+1,k+1)*1/(n+1)
所以
Cn0+1/2Cn1+1/3Cn2+...+1/(n+1)Cnn
=1/(n+1)C(n+1,1)+1/(n+1)C(n+1,2)……+1/(n+1)C(n+1,n+1)
=1/(n+1) [C(n+1,1)+C(n+1,2)……+C(n+1,n+1)]
=1/(n+1)*(2^(n+1)-1)
因为(1+1)^n=C(n,0)+C(n,1)+……+C(n,n)
所以C(n+1,1)+C(n+1,2)……+C(n+1,n+1)=(1+1)^(n+1)-C(n+1,0)=2^(n+1)-1
所以Cn0+1/2Cn1+1/3Cn2+...+1/(n+1)Cnn=1/(n+1)*(2^(n+1)-1)