Question

已知数列{an}的前n项和为Sn，且满足关系an=1/2Sn+1。求：(1)a1a2的值；(2)求数列通项公式an

Answer 1

a1=s1,由an=1/2Sn+1得，a1=0.5a1+1，a1=2a2=0.5(a1+a2)+1,a2=0.5(2+a2)+1,a2=4an-an-1=0.5(sn-sn-1)=0.5an0.5an=an-1,an/an-1=2an=2^n

Answer 2

a1=s1,由an=1/2Sn+1得，a1=0.5a1+1，a1=2a2=0.5(a1+a2)+1,a2=0.5(2+a2)+1,a2=4an-an-1=0.5(sn-sn-1)=0.5an0.5an=an-1,an/an-1=2an=2^n

Answer 3

1/2Sn+1=(1/2)·Sn+1 还是=1/（2Sn+1）？？给个解释

Answer 4

我们学的时候经常碰到这种题，不过根据你这条件只能求出来a1=a2，a3=0，别的什么也求不出来呀。