一道等差数列的题,具体见补充数列{an}中,a1=1,n>=2时,前n项和Sn满足Sn的平方=an（Sn-1）①证明{1/Sn}是等差数列②设bn=log2[Sn/S(n+2)],数列{bn}的前n项和为Tn,求满足Tn>=6的最小正整数n

Question

一道等差数列的题,具体见补充
数列{an}中,a1=1,n>=2时,前n项和Sn满足 Sn的平方=an（Sn-1）
①证明{1/Sn}是等差数列
②设bn=log2[Sn/S(n+2)],数列{bn}的前n项和为Tn,求满足Tn>=6的最小正整数n

Answer 1

(1)由an=S(n-1)可知：
a(n-1)=S(n-2)
两式相减得：an-a(n-1)=a(n-1)
所以：an=2a(n-1)
得出：an/a(n-1)=2
数列{an}是以首相为5,公比为2的等比数列.
an=a1·q^(n-1)=5·2^(n-1)
(2)证明：1/5+1/(5·2)+1/(5·2^2)+……+1/(5·2^(n-1))
=1/5[1+1/2+1/(2^2)+……+1/2^(n-1)]
=1/5[(1-1/2^n)/(1-1/2)]
=2/5·[1-1/2^n]