设正整数数列｛an｝前n项和为Sn,且满足Sn=1/8（an+2）∧2,求证该数列为等差数列

(1)Sn=1/8(an+2)^2,S(n-1)=1/8[a(n-1)+2]^2,
an=Sn-S(n-1)=1/8[an^2+4an-a(n-1)^2-4a(n-1)]
8an=an^2+4an-a(n-1)^2-4a(n-1)
an^2-a(n-1)^2-4an-4a(n-1)=0
[an+a(n-1)][an-a(n-1)]-4[an+a(n-1)]=0
[an+a(n-1)][an-a(n-1)-4]=0
因为{an}为正整数数列,an+a(n-1)>0
所以an-a(n-1)-4=0,即an-a(n-1)=4
所以{an}是等差数列
(2).a1=S1=1/8(a1+2)^2
解方程得:a1=2
根据等差数列求和公式:
数列{an}中,San=na1+n(n-1)d/2=2n+2n(n-1)=2n^2
数列{bn}中,bn=1/2an-30
Sbn=1/2San-30n
带入San=2n^2
Sbn=n^2-30n=(n-15)^2-225
当n=15时,数列{bn}的前n项和最小,为-225.