(1)由题意an=Sn+n① ∴an+1=Sn+1+n+1② ②-①得an+1=an+1, 即an+1+1=a(an+1),{an+1}是以a为公比的等比数列.∴an+1=(a1+1)an-1 又由a1=a1+1⇒a1=a-1∴an=an-1
(2)a=时,bn=n()nlg,bn+1-bn=•()n•lg 当n<8时,bn+1-bn<0即bn+1<bn,∴b1>b2>>b8 当n=8时,bn+1-bn=0即bn+1=b&n,b8=b9 当n>8时,bn+1-bn>0即bn+1>bn∴b9<b10< 存在最小项且第8项和第9项最小
(3)由bn+1>bn得bn+1-bn=(n+1)an+1lga-nanlga=an[(n+1)a-n]lga>0 当a>1时,得(n+1)a-n>0,即a>,显然恒成立,∴a>1 当0<a<1时,lga<0,∴(n+1)a-n<0即a<,∴a<,∴0<a< 综上,a的取值范围为(0,)∪(1,+∞). |