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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...

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)^(...

(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...

(1)∵a1=3,an=2Sn+1+3n(n∈N*,n≥2),∴当n≥2时,an=Sn-Sn-1,∴Sn-3Sn-1=3n,∴Sn3n-Sn?13n?1=1,∴数列{Sn3n}是以1为首项,1为公差的等差数列;(2)由(1)得Sn3n=n,∴Sn=n?3n,∴n≥2时,an=(2n+1)?3n-1,n=1时也成立,∴an=(2n+1)?3n-1;(...

∵数列{an}满足an+1=an-57,且a1=5,∴数列{an}是公差d=-57,首项a1=5的等差数列,∴Sn=5n+n(n?1)2×(?57)=-514n2+7514n=-514(n2-15n)=-514(n-152)2+112556,∴n=7或n=8时,Sn取得最大值.故选:C.

(1)4Sn-4Sn-1=2an+an^2-(2an-1+an-1^2) 4an=2an+an^2-(2an-1+an-1^2) an-an-1=2 所以公差为2 4S1=2a1+a1^2,S1=a1 所以a1=2 所以an=2n+2 亲 记得采纳哦 O(∩_∩)O谢谢

解: a1a2=2(S1+1)=2(a1+1) a1=2代入, 2a2=2(2+1)=6 a2=3 n≥2时, ana(n+1)=2(Sn+1)=2Sn+2 a(n+1)a(n+2)=2S(n+1)+2 [2S(n+1)+2]-(2Sn+2)=2a(n+1)=a(n+1)a(n+2)-ana(n+1) 数列是正项数列,a(n+1)>0,等式两边同除以a(n+1) a(n+2)-an=2,为定值 ...

(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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