Question

已知数列an的各项均为正数,前n项和为Sn且Sn等于二分之an(an+1),n属于正数 。1求数列

已知数列an的各项均为正数,前n项和为Sn且Sn等于二分之an(an+1),n属于正数 。1求数列an的通项公式，2 bn等于2sn分之一，tn等于b1+b2+……+bn.求Tn

Answer 1

Sn=an(an+1)/2 (1)

n=1
2a1=a1(a1+1)
(a1)^2 -a1 =0
a1=1

S(n-1)=a(n-1)(a(n-1)+1)/2 (2)
(1)-(2)
2an =an(an+1) -a(n-1)(a(n-1)+1)
(an)^2-[a(n-1)]^2 - an - a(n-1)=0
[an + a(n-1)] .[an-a(n-1) -1]=0
an-a(n-1) -1=0
an-a1=n-1
an =n
(2)
Sn = n(n+1)/2
bn = 1/(2Sn)
= 1/[n(n+1)]
= 1/n -1/(n+1)
Tn =b1+b2+...+bn
= 1- 1/(n+1)
= n/(n+1)