题目

已知n∈N*，数列{an}和{bn}满足a1=b1=6,a2=4，数列{bn}的前n项和为Sn且Sn+1=Sn+n+6,(1)求证:{bn-2}是等比数列；(2)若数列{an+1-an}(n∈N*)是公差为1的等差数列，试比较an与bn的大小. 答案：解：(1)证明:当n≥2时，Sn+1=Sn+n+6,                   ①Sn=Sn-1+n+5,                                                      ②①-②,得bn+1=bn+1.                                                         ∴bn+1-2=(bn-2)，即n≥2时{bn-2}是等比数列.又S2=S1+1+6,∴b2=-b1+1+6-b1=4.∴b2-2= (b1-2)=2,即n∈N*时{bn-2}是等比数列.                                 (2)由(1)知bn-2=(b1-2)·()n-1,即bn=2+8·()n.                                 由已知a2-a1=-2,∴an+1-an=(a2-a1)+(n-1)·1=n-3.n≥2时，an=(an-an-1)+(an-1-an-2)+…+(a3-a2)+(a2-a1)+a1=(n-4)+(n-5)+…+(-1)+(-2)+6=.n=1也合适.∴an=(n∈N*).                                                    设f(n)=an-bn=n2-n+7-8·()n=(n-)2+-8·()n.                         当n≥4时(n-)2+为n的增函数，-8·()n也为n的增函数，∴当n≥4时有f(n)≥f(4)=，即an-bn≥.                                      又f(1)=f(2)=f(3)=0,∴对n∈N*都有an≥bn.

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