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已知正项数列{an}的前n项和为Sn,且a1=1, a²n+...

解: (1) a2²=S2+S1=a1+a2+a1=2a1+a2=2×1+a2=a2+2 a2²-a2-2=0 (a2+1)(a2-2)=0 a2=-1(舍去)或a2=2 a(n+1)²=S(n+1)+Sn a(n+2)²=S(n+2)+S(n+1) a(n+2)²-a(n+1)²=S(n+2)-Sn=a(n+2)+a(n+1) [a(n+2)+a(n+1)][a(n+2)-a(...

(Ⅰ)解:∵4Sn=an?an+1,n∈N* ①,∴4a1=a1?a2,又a1=2,∴a2=4.当n≥2时,4Sn-1=an-1?an ②,①-②得:4an=an?an+1-an-1?an,由题意知an≠0,∴an+1-an-1=4,当n=2k+1,k∈N*时,a2k+2-a2k=4,即a2,a4,…,a2k是首项为4,公差为4的等差数列,∴a2k=4+4...

(Ⅰ)方法1:由题意得an+1=2Sn+1,an=2Sn-1+1(n≥2)两式相减得an+1-an=2(Sn-Sn-1)=2an.an+1=3an(n≥2)所以当n≥2时,{an}是以3为公比的等比数列.要使n∈N*时,{an}是等比数列,则只需a2a1=2t+1t=3?t=1方法2:由题意,a1=t,a2=2S1+1=2t+1...

等式两边同时除以啊an*a(n+1),1/an-1/a(n+1)=-4,即an的数成等差数列。接下来应该不说了吧。

解答:证明:(1)∵rSn=anan+1-1,①∴rSn+1=an+1an+2-1,②②-①,得:ran+1=an+1(an+2-an),∵an>0,∴an+2-an=r.…(4分)(2)当n=1时,ra=aa2-1,∴a2=1+ara=r+1a,根据数列是隔项成等差,写出数列的前几项:a,r+1a,a+r,2r+1a,a+2r,3r+1a...

1. a(n+1)=(1/3)Sn S(n+1)-Sn=(1/3)Sn S(n+1)=(4/3)Sn S(n+1)/Sn=4/3,为定值。 S1=a1=1 数列{Sn}是以1为首项,4/3为公比的等比数列。 Sn=1×(4/3)^(n-1)=(4/3)^(n-1) n≥2时, an=Sn-S(n-1) =(4/3)^(n-1)-(4/3)^(n-2) =(4/3)×(4/3)^(n-2)-(4/3)^(...

1、s2=a1+a2,s1=a1 所以令n=1 a2=2(√a1)+1=3 2、a(n+1)-1=2√Sn 所以Sn=1/4(a(n+1)-1)² Sn-1=1/4(an-1)² 两式相减 an=1/4【(a(n+1)-1)²-(an-1)²】 4an+(an-1)²=a(n+1)-1)² (a(n+1)-1)²=...

(1)①由题意,得2a1+4d=178a1+28d=56,解得d=-1…(4分)②由①知a1=212,所以an=232?n,则bn=3n?an=3n?(232?n),…(6分)因为bn+1-bn=2×3n×(10-n)…(8分)所以b11=b10,且当n≤10时,数列{bn}单调递增,当n≥11时,数列{bn}单调递减,故当n...

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