（1）2a(n+1)=(1+1/n)^2*an=((n+1)^2/n^2)*an2a(n+1)/(n+1)^2=an/n^22b(n+1)=bnb(n+1)=bn/2得证（2）b1/1^2=1bn=2^(1-n)an=n^2*2^(1-n)=n^2/2^(n-1)Cn=a(n+1)-(1/2)*an=(n+1)^2/2^n-n^2/2^n=(2n+1)/2^nSn可用错位相...