问题
解答题
若Sn是公差不为0的等差数列{an}的前n项和,且S1,S2,S4成等比数列. (1)求等比数列S1,S2,S4的公比; (2)若S2=4,求{an}的通项公式; (3)设bn=
|
答案
(1)∵数列{an}为等差数列,∴S1=a1,S2=a2+d,S4=a4+6d,
∵S1,S2,S4成等比数列,∴S1•S4=S 22
∴a1(4a1+6d)=(2a1+d)2,∴2a1d=d2
∵公差为d不等于0,∴d=2a1,
∴q=
=S2 S1
=4,4a1 a1
(2)∵S2=4,∴2a1+d=4,
∵d=2a1,∴a1=1,d=2,
∴an=2n-1
(3)∵bn=
=3 (2n-1)(2n+1)
(3 2
-1 2n-1
)1 2n+1
∴Tn=
[(1-3 2
)+(1 3
-1 3
)+…+(1 5
-1 2n-1
)]=1 2n+1
(1-3 2
)1 2n+1
∴(Tn)min=1
使得Tn>
对所有n∈N*都成立,等价于1>m 20
,∴m<20m 20
∴m的最大值为19.