已知数列{an}中,a1=1,an+1(2+an)=2an(n∈N*),(Ⅰ)求a2,a3,a4的值及数列{an}的通项公式;(Ⅱ)记Tn=a1a2+a2a3+…+an-1an(n≥2),试判断Tn与

题目简介

已知数列{an}中,a1=1,an+1(2+an)=2an(n∈N*),(Ⅰ)求a2,a3,a4的值及数列{an}的通项公式;(Ⅱ)记Tn=a1a2+a2a3+…+an-1an(n≥2),试判断Tn与

题目详情

已知数列{an}中,a1=1,an+1(2+an)=2an(n∈N*),
(Ⅰ)求a2,a3,a4的值及数列{an}的通项公式;
(Ⅱ)记Tn=a1a2+a2a3+…+an-1an(n≥2),试判断Tn与2的大小,并说明理由.
题型:解答题难度:中档来源:不详

答案

(Ⅰ)由an+1(2+an)=2an(n∈N*),得an+1=
2an
2+an

∵a1=1,∴a2=
2a1
2+a1
=class="stub"2
3
a2=
2a2
2+a2
=class="stub"2
4
a4=
2a3
2+a3
=class="stub"2
5
.  …(3分)
又由an+1=
2an
2+an
class="stub"1
an+1
=class="stub"1
an
+class="stub"1
2
,即class="stub"1
an+1
-class="stub"1
an
=class="stub"1
2

∴{class="stub"1
an
}是以1为首项,class="stub"1
2
为公差的等差数列,
class="stub"1
an
=1+class="stub"1
2
(n-1)=class="stub"n+1
2
,∴an=class="stub"2
n+1
.                                    …(7分)
(Ⅱ)Tn<2. 证明如下:…(8分)
当n≥2时,an-1an=class="stub"2
n
class="stub"2
n+1
=4(class="stub"1
n
-class="stub"1
n+1
),…(10分)
∴Tn=4[(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
)]=4(class="stub"1
2
-class="stub"1
n+1
)=2-class="stub"4
n+1
<2…(15分)

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