Question

已知数列{an}中,前n项和为Sn,an>0,且2√Sn=an+1,求an

Answer 1

a1=s1解得a1=1;

sn=(an+1)^2/4,sn-1=(an-1)^2/4;

an=sn-sn-1 得(an+an-1)(an-an-1-2)=0

an>0,an+an-1不等于0

故an-an-1=2=d,又因为a1=1

所以an=2n-1.