Question

已知数列{an}中a1=1,an+1=an/2an+1(n∈N+).(1)求证数列{1/an}为等差数列

（2）设bn=an*an+1(n∈N+)，数列{bn}的前n项和为Sn，求满足Sn>1005/2012的最小正整数n。

Answer 1

(1)
a(n+1)=an/(2an+1)
1/a(n+1) = 2 + 1/an
1/a(n+1) - 1/an =2
=>{1/an}为等差数列
(2)
1/a(n+1) - 1/an =2
1/an -1/a1 = 2(n-1)
an = 1/(2n-1)
bn = an.a(n+1)
= 1/[(2n-1)(2n+1)]
= (1/2)[1/(2n-1) -1/(2n+1)]
Sn = b1+b2+...+bn
= (1/2)[ 1/1 - 1/(2n+1) ]
= n/(2n+1) >1005/2012
2012n >1005(2n+1)
2n > 1005
n > 1005/2
最小正整数n= 503