数列bn的首项b1=1其前n项和b=1/2(n+1)bn 求通项公式

题目

数列bn的首项b1=1其前n项和b=1/2(n+1)bn 求通项公式

答案

前n项和Sn = （n+1）bn /2
前n-1项和Sn-1 = (n bn-1)/2
则bn = Sn - (Sn-1) = [(n+1)bn - (n bn-1)]/2
整理得：bn = [n/ (n-1)] bn-1
= [n/(n-1) ] * [(n-1)/(n-2)] * bn-2
= [n/(n-2)] *bn-2
= [n/(n-2)] *[(n-2/(n-3))] *bn-3
= [n/(n-3)] *bn-3
:
:
:
= (n/2) *b2
=nb1 = n