问题 解答题
已知数列{an}满足a1=4,an+1=an+p•3n+1(n∈N*,p为常数),a1,a2+6,a3成等差数列.
(Ⅰ)求p的值及数列{an}的通项公式;
(Ⅱ)设数列{bn}满足bn=
n2
an-n
,证明:bn
4
9
答案

(Ⅰ)因为a1=4,an+1=an+p•3n+1

所以a2=a1+p•31+1=3p+5a3=a2+p•32+1=12p+6

因为a1,a2+6,a3成等差数列,所以2(a2+6)=a1+a3

即6p+10+12=4+12p+6,所以p=2.

依题意,an+1=an+2•3n+1

所以当n≥2时,a2-a1=2•31+1a3-a2=2•32+1

an-1-an-2=2•3n-2+1an-an-1=2•3n-1+1

相加得an-a1=2(3n-1+3n-2+…+32+3)+n-1

所以an-a1=2

3(1-3n-1)
1-3
+(n-1),

所以an=3n+n

当n=1时,a1=31+1=4成立,

所以an=3n+n.                            

(Ⅱ)证明:因为an=3n+n,所以bn=

n2
(3n+n)-n
=
n2
3 n

因为bn+1-bn=

(n+1)2
3n+1
-
n2
3n
=
-2n2+2n+1
3n+1
,(n∈N*).

若-2n2+2n+1<0,则n>

1+
3
2
,即n≥2时,bn+1<bn

又因为b1=

1
3
b2=
4
9
,所以bn
4
9

单项选择题 A1/A2型题
单项选择题