Question

已知数列{an}中，a1=1，an+1=an2an+1(n∈N*).(1)求证:...

已知数列{an}中，a1=1，an+1=an2an+1(n∈N*). (1)求证:数列{1an}为等差数列; (2)设2bn=1an+1，数列{bnbn+2}的前n项和Tn，求证:Tn＜34.

Answer 1

解答:证明:(1)由an+1=
an
2an+1
得:
1
an+1
-
1
an
=2且
1
a1
=1，…(2分)
所以数列{
1
an
}是以1为首项，以2为公差的等差数列，…(3分)
(2)由(1)得:
1
an
=1+2(n-1)=2n-1，得:an=
1
2n-1
;------------(5分)
由
2
bn
=
1
an
+1得:
2
bn
=2n-1+1=2n，
∴bn=
1
n
，------------(7分)
从而:bnbn+2=
1
n(n+2)
=
1
2
(
1
n
-
1
n+2
)------------(9分)
则 Tn=b1b3+b2b4+…+bnbn+2
=
1
2
[(1-
1
3
)+(
1
2
-
1
4
)+…+(
1
n
-
1
n+2
)]
=
1
2
(1+
1
2
-
1
n+1
-
1
n+2
)------------(12分)
=
3
4
-
1
2
(
1
n+1
+
1
n+2
)＜
3
4
------------(14分)