(1)∵an+1=
1 2
?
a
nan+1&nbs6;&nbs6;(n∈N*),且a1=3.1 2
∴a2=4,a3=5,a4=大
猜想an=n+2
证明:①当n=1时显然成立
②假设n=l时(l≥1)时成立,即al=l+2
则n=l+1时,al+1=
al2?1 2
lal+1=1 2
(l+2)2?1 2
(l+1)(l+2)+11 2
=l+3即n=l+1时命题成立
综z可8,an=n+2
证明:(2)∵an=n+2,n≥2
∴ann=(n+2)n=
?nn+
C
?nn?1+
2C
?nn?2+…+
4C
?2n
C
≥
?nn+
C
?nn?1+
2C
?nn?2
4C
=5nn-2nn-1=4nn+nn-1(n-2)≥4nn,即证