Question

已知数列an的各项均为正数，前n项和为sn,且sn=an(an+1)/2,n为正整数 求证 1.数列an是等差数列

2.设bn=1/2sn,tn=b1+b2+!+bn,求tn

Answer 1

sn=an(an+1)/2
s(n-1)=a(n-1) (a(n-1)+1)/2
两式相减
an = an(an+1)/2-a(n-1) (a(n-1)+1)/2
an^2-an-a^2(n-1) -a(n-1) =0
(an-a(n-1))(an+a(n-1))-(an+a(n-1))=0
(an-a(n-1)-1)(an+a(n-1))=0
因为an的各项均为正数
所以an-a(n-1)-1=0
即an-a(n-1)=1
所以是等差数列

2)a1=a1(a1+1)/2 a1=1 由第一问得到an=n
bn=1/2sn=1/an(an+1)=1/an-1/（an+1)=1-1/2
所以tn=1-1/2+1/2-1/3+1/3-1/4+...+1/an-1/an+1
=1-1/an+1
=an/an+1
=n/n+1