Question

如果数列{an}的前n项和满足2Sn=an+2/an,则数列{Sn的平方}的通项公式是

Answer 1

n=1时,可得：2a1=a1+2/a1,即：S1^2=a1^2=2；
对2Sn=an+2/an,和2S(n-1)=a(n-1)+2/a(n-1),
做差,得：2an=an-a(n-1)+2/an-2/a(n-1),
即：an+a(n-1)=2/an-2/a(n-1)；
又有：Sn^2-S(n-1)^2=an^2+2anS(n-1)
=an^2+an(a(n-1)+2/a(n-1))
=an(an+a(n-1))+2an/a(n-1)
=an(2/an-2/a(n-1))+2an/a(n-1)
=2+2an/a(n-1)-2an/a(n-1)
=2,
综上,故而Sn^2=2n.