广东省2013届高三最新理科试题精选（37套含13大市区的二模）分类(6)

(3) 2a1?f(2)?3,得

a1?3?122,又an?1?f(an)?an?an?1.

?an?1?an?an2?2an?1?(an?1)2?0,

?an?1?an?1

由an?1??1?an2?an?1,得an?1?1=an(an?1),

1nnan?1?1a(a?1)a?1an?1?1n1,即a12?1nna?1a1?1n?1?1

11?S?1111aa??12???1a?(112013a?1a???1)?(a2?1?1a3?1)???(a2013?1?a2014?1

)?12014a?1a?1?2?1a2014?12

又

S?11aa?12?2437??2621?1 [来源:学,科,网Z,X,X,K]

即1?S?2,故S的整数部分为. l4分