已知数列{an}中,a1=1,an+1=an2an+1(n∈N).(1)求数列{an}的通项公式an;(2)设:2bn=1an+1求数列{bnbn+1}的前n项的和Tn;(3)已知P=(1+b1)(1

题目简介

已知数列{an}中,a1=1,an+1=an2an+1(n∈N).(1)求数列{an}的通项公式an;(2)设:2bn=1an+1求数列{bnbn+1}的前n项的和Tn;(3)已知P=(1+b1)(1

题目详情

已知数列{an}中,a1=1,an+1=
an
2an+1
(n∈N).
(1)求数列{an}的通项公式an
(2)设:
2
bn
=
1
an
+1
 求数列{bnbn+1}的前n项的和Tn
(3)已知P=(1+b1)(1+b3)(1+b5)…(1+b2n-1),求证:Pn>
2n+1
题型:解答题难度:中档来源:不详

答案

(1)由an+1=
an
2an+1
得:class="stub"1
an+1
-class="stub"1
an
=2
class="stub"1
a1
=1

所以知:数列{class="stub"1
an
}是以1为首项,以2为公差的等差数列,
所以 class="stub"1
an
=1+2(n-1)=2n-1
,得an=class="stub"1
2n-1

(2)由class="stub"2
bn
=class="stub"1
an
+1
得:class="stub"2
bn
=2n-1+1=2n
,∴bn=class="stub"1
n

从而:bnbn+1=class="stub"1
n(n+1)

则 Tn=b1b2+b2b3+…+bnbn+1=class="stub"1
1×2
+class="stub"1
2×3
+…+class="stub"1
n(n+1)

=(1-class="stub"1
2
)+(class="stub"1
2
-class="stub"1
3
)+(class="stub"1
3
-class="stub"1
4
)+…+(class="stub"1
n
-class="stub"1
n+1

=1-class="stub"1
n+1
=class="stub"n
n+1

(3)已知Pn=(1+b1)(1+b3)(1+b5)…(1+b2n-1)=class="stub"2
1
×class="stub"4
3
×class="stub"6
5
×…×class="stub"2n
2n-1

∵(4n)2<(4n)2-1,∴class="stub"2n+1
2n
<class="stub"2n
2n-1

设:Tn=class="stub"3
2
×class="stub"5
4
×…×class="stub"2n+1
2n
,则Pn>Tn
从而:Pn2PnTn=class="stub"2
1
×class="stub"3
2
×class="stub"4
3
×…×
class="stub"2n
2n-1
×class="stub"2n+1
2n
=2n+1

故:Pn>
2n+1

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