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⇒
=2,又∵a1=2an+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
(1-1 2
)1 2n-1
要使
(1-1 2
)<1 2n-1
所有的n∈N*都成立,必须且仅需满足m 30
≤1 2
⇒m≥15m 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
Tn=1 2
+1 22
+3 23
+…+5 24 2n-1 2n+1
相减得:
Tn=1 2
+(1 2
+1 2
+…+1 22
)-1 2n-1 2n-1 2n+1
化简得Tn=3-
-1 2n-2
<32n-1 2n
所以Tn<3