Ⅰ、a2=3a1=9/2
a3=a2+2=13/2
a4=3a3=39/2
Ⅱ、bn=a﹙2n-1）+n+1/2
b1=a1+1+1/2=3
2n+1为奇数
b(n+1)=a﹙2n+1）+n+1+1/2
=a(2n)+2n+n+1+1/2
=3a﹙2n-1）+3n+3/2
=3(a﹙2n-1）+n+1/2)
=3bn
所以数列﹛bn﹜是一个首项是3,公比是3的等比数列
Ⅲ、由上例可以看出a﹙2n-1）=bn-n-1/2
又a﹙2n)=3a﹙2n-1）
所以Sn=4{[3(1-3^N)/(1-3)]-n(n+1)/2-1/2n}
=6(3^N-1)-2n^2-4n
S2n+3=6(3^2N-1)-8n^2-8n+3=6*3^2N-8n^2-8n-3
S2（n+1）+3=6*3^(2N+1)-2(2n+1)^2-4(2n+1)-3=6*3^(2N+1)-8n^2-16n-12
S2（n+1）+3-3 (S2n+3)=16n^2+8n-3>0(n∈正整数)
S2（n+1）+3>3 (S2n+3)
[S2（n+1）+3]／3^(n+1)>[S（2n)+3]/3^n
累死我了,