∴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...
∴(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_...
2an、2Sn、an²成等差数列，则 2·2Sn=2an+an²4Sn=2an+an²令n=1，得 4S1=4a1=2a1+a1²a1²-2a1=0 a1(a1-2)=0 a1=0(舍去)或a1=2 a1=2 (2)n≥2时，4Sn-4S(n-1)=4an=2an+an²-[2a(n-1)+a(n-1)²]an²-a(n-1)&...
∴{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...
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...
解答:(1)证明:由题意可知2Sn=nan+n 当n≥2时,2Sn-1=(n-1)an-1+n-1 相减得2an=nan-(n-1)an-1+1 即(n-2)an-(n-1)an-1+1=0 ① 所以(n-3)an-1-(n-2)an-2+1=0 ② 由①-②得(n-2)an-2(n-2)an-1+(n-2)an-2=0(n≥3) ...
证明:(1)∵2Sn-nan=n,∴2Sn+1-(n+1)an+1=n+1,相减可得:nan-(n-1)an+1=1,n≥2时,(n-1)an-1-(n-2)an=1,相减可得:2an=an-1+an+1,∴数列{an}是等差数列.(2)由题意可得:当n=1时,2S1-a1=1,解得a1=1.故Sn+1=(n+1)+d,∴=,若an+1≤bn<an+2;∴1+nd≤<1+(n+1)d...
∴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),...
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 ...