∴(an+an-1)(an-an-1-1)=0,∵an+an-1>0,∴an-an-1=1则数列{an}是以1为首项,1为公差的等差数列∴an=a1+(n-1)d=1+(n-1)=n.(Ⅱ)∵ bn= n 2n,∴ Tn= 1 2+ 2 22+ 3 23+…+ n 2n①又因为 1 2Tn= 1 22+ 2 23+ 3 24+…+ n 2n+1...
解答 解:由2Sn=nan+2,得2S2=2(a1+a2)=2a2+2,解得a1=1;当n≥2时,有2Sn-1=(n-1)an-1+2,两式作差可得:2an=nan-(n-1)an-1(n≥2),即(n-2)an=(n-1)an-1,∴当n≥3时,有(a_n)/(a_(n-1))=(n-1)/(n-2),∴(a_3)/(a_2)=2/1,(a_4)/(a_3)=3/2,(a_5)/(a_...
∴an+an?1=an2?an?12,∴an-an-1=1,∴{an}是公差为1的等差数列,由2S1=a12+a1,得a1=1,∴an=1+(n-1)×1=n.(2)bn=2anlog 1 22an=-n?2n,∴Hn=-(1×2+2×22+3×23+…+n×2n),∴2Hn=-(22+2×23+3×24+…+n×2n+1),...
∴an+1-1=an,或an+1-1=-an, 即an+1-an=1或an+1+an=1 若an+1-an=1 即数列{an}为等差数列,公差d=1. 若an+1+an=1 ∵当n=1时,2S1=a12+1-4, ∴2a1=a12-3, 解得a1=-1(舍)或a1=3, 当a1=3时,a2=1-a1=1-3=-2不满足条件. (2)∵2Sn=an2+n-4, ∴当n=1时,2a1=a12...
解答: 解:(I)当n=1时,a1=S1>0,所以2a1=a12+a1⇒a1=1,又2Sn=an2+an,2Sn+1=an+12+an+1,两式相减得:an+1-an=1,所以数列{an}是首项为1,公差为1的等差数列故数列{an}的通项公式为an=n…(6分)(II)由(Ⅰ)得 bn= 2n (2n-1) (2n+1-1)+(-1)nn,记数列{bn}的前2n项和为T2n,则 ...
-2an-2a(n-1)=0 [an+a(n-1)][an-a(n-1)-2]=0 数列各项均为正，an+a(n-1)恒>0，因此只有an-a(n-1)-2=0 an-a(n-1)=2，为定值 数列{an}是以2为首项，2为公差的等差数列 an=2+2(n-1)=2n n=1时，a1=2×1=2，同样满足表达式 数列{an}的通项公式为an=2n ...
即an-an-1=1,(n∈N*,n⩾2)所以数列{an}是等差数列.(2)由(1)可得an=2+n-1=n+1,n∈N*,则b_n=((-1))^na_n^2=((-1))^n((n+1))^2,n∈N*,从而b_(2n-1)+b_(2n)=-(2n)^2+(2n+1)^2=4n+1,故T2n=b1+b2+…+b2n-1+b2n(4+1)+(4×2+1)+…+(4n+1)=((5...
∴{an}是以1为首项,以1为公差的等差数列.∴an=n,2Sn=n2+n.∴cn=(-1)n • 2n+1 n2+n=(-1)n( 1 n+ 1 n+1).设cn的前n项和为Tn,则T2017=-1- 1 2+ 1 2+ 1 3- 1 3 - 1 4+…- 1 2017- 1 2018=-1- 1 2018=- 2019 2018.故答案为: - 2019 2018. 利用an=Sn-Sn-1判断...
【解答】解:(I)当n=1时,a1=S1>0,所以2a1=a12+a1⇒a1=1,又2Sn=an2+an,2Sn+1=an+12+an+1,两式相减得:an+1-an=1,所以数列{an}是首项为1,公差为1的等差数列故数列{an}的通项公式为an=n…(6分)(II)由(Ⅰ)得bn=2n(2n-1) (2n+1-1)+(-1)nn,记数列{bn}的前2n项和为T2n,则T...
分析:(1)由已知条件得2an=2Sn-2Sn-1=2n,从而得到an=n(n≥2),又n=1时,a1=1适合上式.由此能求出数列{an}的通项公式. (2)bn= 1 anan+1 +2an-1=( 1 n - 1 n+1 )+(2n-1),由此能求出数列{bn}的前n项和Sn. 解答:解:(1)∵数列{an}的前n项和为Sn,且2Sn=n2+n, ...