Question

已知数列{an}中，an>0,Sn=a1+a2+……+an, 且an=6Sn/（an+3) 求Sn

Answer 1

an=6Sn/（an+3)

an^2+3an=6Sn 1)

a(n-1)^2+3a(n-1)=6S(n-1) 2)

an=Sn-S(n-1) 3)

1)-2)再代入3)整理有

(An-A(n-1))*(An+A(n-1))=3*(An+A(n-1)) 4)

An>0
An+A(n-1)>0

由4)得到
An-A(n-1)=3

3为常数,所以An为等差数列

A1^2+3*A1=6S1=6A1
A1=3

An=A1+(n-1)*d=3+(n-1)*3=3n

Sn=A1+A2+...+An=3*(1+2+3+...+n)=3*(1+n)*n/2

Answer 2

由题意知：an=6Sn/（an+3)
an^2+3an=6Sn
a(n-1)^2+3a(n-1)=6S(n-1)
6an=6Sn-6S(n-1)
(an-a(n-1))*(an+a(n-1))=3*(an+a(n-1))
an>0
an+a(n-1)>0
因此
an-a(n-1)=3
3为常数,所以an为等差数列
a1^2+3*a1=6S1=6a1
a1=3
an=a1+(n-1)*d=3+(n-1)*3=3n
Sn=a1+a2+...+an=3*(1+2+3+...+n)=3*(1+n)*n/2