已知数列{an}的前n项和为Sn,且a1=1,a2=3,2Sn-(n+1)an=An+B(其中A,B是常数,n属于正整数

⑴令n=1,2*S1-2*a1=A+B ==> A+B=0令n=2,2*S2-3*a2=2A+B ==> 2A+B=-1所以A=-1,B=1⑵证明：由⑴,则2Sn-(n+1)an=1-n 。①2*S(n+1)-(n+2)*a(n+1)=-n 。②②-①,整理后得：n*a(n+1)-(n+1)an=1等号两边同除以n(n+1),得：a(n+1)/(n+1)-an/n=1/[n(n+1)]=1/n-1/(n+1)a(n+1)/(n+1)+1/(n+1)-(an/n+1/n)=0所以{an/n+1/n}是等差数列且为常数列。用a1代,得到an/n+1/n=2,所以an=2n-1综上,数列{an/n+1/n}是等差数列,且an=2n-1