在数列{an}中,a1=-13,n∈N*,当n≥2时,有3an-2an-1+n+2=0,设bn=an+n+1.(I)求b1,b2;(II)证明数列{bn-1}是等比数列;(III)设cn=(23)n2

题目简介

在数列{an}中,a1=-13,n∈N*,当n≥2时,有3an-2an-1+n+2=0,设bn=an+n+1.(I)求b1,b2;(II)证明数列{bn-1}是等比数列;(III)设cn=(23)n2

题目详情

在数列{an}中,a1=-
1
3
,n∈N*
,当n≥2时,有3an-2an-1+n+2=0,设bn=an+n+1.
(I)求b1,b2
(II)证明数列{bn-1}是等比数列;
(III)设cn=
(
2
3
)
n
2
b2n
+bn
,求数列{cn}的前n项和Tn
题型:解答题难度:中档来源:成都一模

答案

(I)∵a1=-class="stub"1
3
,bn=an+n+1∴b1=a1+2=class="stub"5
3

当n=2时,3a2-2a1+4=0可得a2=-class="stub"14
9

b2=3+a2=class="stub"13
9

(II)由3an-2an-1+n+2=0得,3(an+n)=2(an-1+n-1)
an+n
an-1+n-1
=class="stub"2
3
,n≥2即
bn-1
bn-1-1
=class="stub"2
3

b1- 1=class="stub"2
3
≠0

{bn-1}是以class="stub"2
3
为首项,class="stub"2
3
为公比的等比数列

(III)由(I)可得bn=class="stub"2
3
bn-1+class="stub"1
3

∴2bn-1+1=3bn,所以bn=1+(class="stub"2
3
)
n

cn=
(class="stub"2
3
)
n
2
b2n
+bn
=
(class="stub"2
3
)
n
(2bn+1)bn
=
bn-bn+1
bnbn+1
=class="stub"1
bn+1
-class="stub"1
bn

Tn=C1+C2+…+Cn=class="stub"1
bn+1
-class="stub"1
b1
=
3n+1
2n+13n+1
-class="stub"3
5

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