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=
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答案
(Ⅰ)因为a1=4,an+1=an+p•3n+1,
所以a2=a1+p•31+1=3p+5;a3=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+1,a3-a2=2•32+1,
…an-1-an-2=2•3n-2+1,an-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
+(n-1),3(1-3n-1) 1-3
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
,(n∈N*).-2n2+2n+1 3n+1
若-2n2+2n+1<0,则n>
,即n≥2时,bn+1<bn.1+ 3 2
又因为b1=
,b2=1 3
,所以bn≤4 9
.4 9