设等比数列{an}的前n项和为Sn,已知an+1=2Sn+2(n∈N*)(I)求数列{an}的通项公式;(Ⅱ)在an与an+1之间插人n个数,使这n+2个数组成公差为dn的等差数列,求数列{1dn}的

题目简介

设等比数列{an}的前n项和为Sn,已知an+1=2Sn+2(n∈N*)(I)求数列{an}的通项公式;(Ⅱ)在an与an+1之间插人n个数,使这n+2个数组成公差为dn的等差数列,求数列{1dn}的

题目详情

设等比数列{an}的前n项和为Sn,已知an+1=2Sn +2(n∈N*)
(I)求数列{an}的通项公式;
(Ⅱ)在an与an+1之间插人n个数,使这n+2个数组成公差为dn的等差数列,求数列{
1
dn
}的前n项和Tn
题型:解答题难度:中档来源:不详

答案

(I)由an+1=2Sn +2(n∈N*)可得an=2sn-1+2(n≥2)
两式相减可得,an+1-an=2an
即an+1=3an(n≥2)
又∵a2=2a1+2,且数列{an}为等比数列
∴a2=3a1
则2a1+2=3a1
∴a1=2
an=2•3n-1
(II)由(I)知,an=2•3n-1an+1=2•3n
∵an+1=an+(n+1)dn
dn=
an+1-an
n+1
=
3n-1
n+1

Tn=class="stub"2
4•30
+class="stub"3
4•31
+class="stub"4
4•32
+…+class="stub"n+1
4•3n-1

class="stub"1
3
Tn
=class="stub"2
4•31
+class="stub"3
4•32
+…+class="stub"n
4•3n-1
+class="stub"n+1
4•3n

两式相减可得,class="stub"2
3
Tn
=class="stub"2
4•30
+class="stub"1
4•3
+class="stub"1
4•32
+…+class="stub"1
4•3n-1
-class="stub"n+1
4•3n

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

=class="stub"5
8
-class="stub"2n+5
8•3n

Tn=class="stub"15
16
-class="stub"2n+5
16•3n-1

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