a_(n1)=f(a_n)=(a+a_n)/(a+a_n),两边取倒数得,1/a_(n+1)=1/a_n+1/a,1/a_1=1,从而{1/a_n}是以1为首项,1/a为等差的等差数列,其通项为(1/a_n)=[(n-1)/a]+1,从而a_n=a/(n+a-1);(2)、可能条件应为bn=a_n-a_(n+1),据此条件,则bn=a_n-a_(n+1)=a/(n+a-1)-a/(n+a),则Sn=[a/a-a/(a+1)]+[a/(a+1)-a/(a+2)]+…+[a/(a+n-1)-a/(a+n)]=1+a/(a+n)