1、当n为奇数时,a（n+2）=an+1,a1＝1,a3＝2,a5＝a3＋1＝3,an＝（n＋1）/2
则对于任意正整数有a（2n－1）＝n
当n为偶数时,a（n+2）=2an,a2＝2,a4＝4,a6＝8,an＝2*2^(n/2-1)
则对于任意正整数有a（2n）＝2*2^(n-1)=2^n
2、bn=a(2n-1)\a2n＝n/[2^n]
sn=1/2+2/4+3/8+ n/[2^n]
0.5sn=(1/2+2/4+3/8+ n/[2^n])*0.5=1/4+2/8+3/16+ n/[2^(n-1)]
sn-0.5sn=1/2-1/4+2/4-2/8+3/8-3/16+ n/[2^n]-n/[2^(n-1)]
1.5sn=1/2+1/4+1/8+1/(2^n)=