问题
解答题
数列{an}的前n项和为Sn,且满足Sn=2(an-1),数列{bn}中,b1=1,且点P(bn,bn+1)在直线x-y+2=0上, (1)求数列{an}、{bn}的通项公式; (2)设Hn=
(3)设Tn=
|
答案
(1)∵Sn=2(an-1),∴Sn+1=2(an+1-1)
两式相减得:an+1=2an+1-2an⇒
an+1 |
an |
∴{an}是以2为首项,以2为公比的等比数列,∴an=2n
又P(bn,bn+1)在直线x-y+2=0上,
∴bn-bn+1+2=0⇒bn+1-bn=2,
又∵b1=1,∴}、{bn}是以1为首项,以2为公差的等差数列,∴bn=2n-1
(2)
1 |
bn-1bn |
1 |
(2n-3)(2n-1) |
1 |
2 |
1 |
2n-3 |
1 |
2n-1 |
∴Hn=
1 |
b1b2 |
1 |
b2b3 |
1 |
bn-1bn |
1 |
2 |
1 |
2n-1 |
要使
1 |
2 |
1 |
2n-1 |
m |
30 |
1 |
2 |
m |
30 |
所以满足要求的最小正整数为15,
(3)Tn=
1 |
2 |
3 |
22 |
5 |
23 |
2n-1 |
2n |
1 |
2 |
1 |
22 |
3 |
23 |
5 |
24 |
2n-1 |
2n+1 |
相减得:
1 |
2 |
1 |
2 |
1 |
2 |
1 |
22 |
1 |
2n-1 |
2n-1 |
2n+1 |
化简得Tn=3-
1 |
2n-2 |
2n-1 |
2n |
所以Tn<3