数列{an}前n项和为Sn,对一切正整数n都有Sn=n+(1/2)an,求an,Sn

因为Sn=n+(1/2)an,所以Sn+1=n+1+(1/2)an+1,两式相减得an+1=1+(1/2)an+1-(1/2)an,所以整理后可得an+1=-an+2,当n=1时,a1=2,所以a2=0,a3=2,a4=0……,所以数列an是一个循环数列,列分段形式,an=2(n=2k+1,k属于整数）an+0(...