问题
解答题
已知数列{an}中a1=1,an+1=
(1)求证:数列{
(2)设bn=an•an+1(n∈N+),数列{bn}的前n项和为Sn,求满足Sn>
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答案
(1)证明:由a1=1与an+1=
得an≠0,an 2an+1
=1 an+1
=2+2an+1 an
,1 an
所以对∀n∈N+,
-1 an+1
=2为常数,1 an
故{
}为等差数列;1 an
(2)由(1)得
=1 an
+2(n-1)=2n-1,1 a1
bn=an•an+1=
=1 (2n-1)(2n+1)
(1 2
-1 2n-1
),1 2n+1
所以Sn=b1+b2+…+bn=
(1-1 2
)+1 3
(1 2
-1 3
)+…+1 5
(1 2
-1 2n-1
)=1 2n+1
(1-1 2
)=1 2n+1
,n 2n+1
由Sn>
即1005 2012
>n 2n+1
,得n>1005 2012
=5021005 2
,1 2
所以满足Sn>
的最小正整数n=503.1005 2012