取倒数得：1/[a(n＋1)]=(1/2)＋1/[an],
1/[a(n＋1)]－1/[an]=1/2,
{1/(an)}是以1/a1为首项,以d=1/2为公差的等差数列.1/(an)=1/(a1)＋(1/2)(n－1),以n=2011代入,算出a1再代入得1/an,再得an.