已知数列{an}中,a1=2,a2=4,an+1=3an-2an-1(n≥2,n∈N*). (Ⅰ)证明数列{an+1-an}是等比数列,并求出数列{an}的通项公式; (Ⅱ)记bn=2(an−1)an,数列{bn}的前n项和为Sn,求使Sn

问题描述:

已知数列{an}中,a1=2,a2=4,an+1=3an-2an-1(n≥2,n∈N*).
(Ⅰ)证明数列{an+1-an}是等比数列,并求出数列{an}的通项公式;
(Ⅱ)记bn

2(an−1)
an
,数列{bn}的前n项和为Sn,求使Sn>2010的n的最小值.

(I)∵an+1=3an-2an-1(n≥2)
∴(an+1-an)=2(an-an-1)(n≥2)
∵a1=2,a2=4∴a2-a1=2≠0,∴an+1-an≠0
故数列{an+1-an}是公比为2的等比数列
∴an+1-an=(a2-a1)2n-1=2n
∴an=(an-an-1)+(an-1-an-2)+(an-2-an-3)++(a2-a1)+a1
=2n-1+2n-2+2n-3++21+2
=

2(1−2n−1)
1−2
+2=2n(n≥2)
又a1=2满足上式,
∴an=2n(n∈N*
(II)由(I)知bn
2(an−1)
an
=2(1−
1
an
)
=2(1−
1
2n
)=2−
1
2n−1

Sn=2n−(1+
1
21
+
1
22
++
1
2n−1
)

=2n−
1−
1
2n
1−
1
2

=2n−2(1−
1
2n
)

=2n−2+
1
2n−1

由Sn>2010得:2n−2+
1
2n−1
>2010

n+
1
2n
>1006
,因为n为正整数,所以n的最小值为1006