已知数列an中,Sn为前n项和,且an>0,且4Sn=an² 2an 1
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1.Sn=-2an+3有S(n-1)=-2a(n-1)+3则an=Sn-S(n-1)=-2an+2a(n-1)=>an=a(n-1)*2/3所以,{an}为共比数列,q=2/32.Sn=-2an+3有
因为An+1=2SnAn=2S(n-1)所以A(n+1)-An=2AnA(n+1)/An=3是公比为3,首项a1=1的等比数列,An=A1*q^(n-1)即An=3^(n-1)
(1)数列{an}中,a1=1,前n项和Sn=n+23an,可知S2=43a2,得3(a1+a2)=4a2,解得a2=3a1=3,由S3=53a3,得3(a1+a2+a3)=5a3,解得a3=32(a
(1)由已知,n,an,Sn成等差数列,所以Sn=2an-n,Sn-1=2an-1-(n-1),(n≥2)两式相减得an=Sn-Sn-1=2an-2an-1-1,即an=2an-1+1,两边加上1,得
设公比为q,当q=-1时,等比数列{an}的各项是a,-a,a,-a,a,-a…的形式,a≠0.又已知Sn是实数等比数列{an}前n项和,故当n为偶数时,Sn=0,当n为奇数时,Sn=a,故选D.
因为Sn=n^2*an.1Sn-1=(n-1)^2*an-1n≥2.21-2:an=n^2*an-(n-1)^2*an-1(n^2-1)*an=(n-1)^2*an-1(n+1)*an=(n-1)*a
Sn=n-5an-85(1)S(n+1)=n+1-5a(n+1)-85(2)(2)-(1)整理得6a(n+1)=1+5an即a(n+1)-1=(5/6)(an-1)又由S1=a1=1-5a1-85得a
因为:An+1=2Sn,则A(n-1)+1=2S(n-1)那么:2Sn-2S(n-1)=(An+1)-(A(n-1)+1)(n>=2)又因为:2Sn-2S(n-1)=2An(n>=2)所以:2An=(
(2n+1)^2-(2n-1)^2=4n^2+4n+1-(4n^2-4n+1)=8nAn=[(2n+1)^2-(2n-1)^2]/[(2n-1)^2(2n+1)^2]=(2n+1)^2/[(2n-1)
n=1时,a1=S1=k+2n≥2时,Sn=2n²+kS(n-1)=2(n-1)²+kan=Sn-S(n-1)=2n²+k-2(n-1)²-k=4n-2数列{a
(1)证明:∵Sn=n-5an-85,n∈N*(1)∴Sn+1=(n+1)-5an+1-85(2),由(2)-(1)可得:an+1=1-5(an+1-an),即:an+1-1=56(an-1),从而{
S1=a1=1-1*a12a1=1a1=1/2S2=1-2a2=a1+a2=1/2+a23a2=1/2a2=1/6Sn=1-nanSn-1=1-(n-1)a(n-1)相减an=Sn-Sn-1=1-na
Sn=n(an+1)/2S(n+1)=(n+1)[a(n+1)+1]/2用下式减上式a(n+1)=[(n+1)a(n+1)-nan+1]/2即2a(n+1)=[(n+1)a(n+1)-nan+1]即(
当n=1时,S1=a1=1/2(a1^2+a1),解得a1=1当n>1时,an=Sn-S(n-1)=1/2(an^2+an)-1/2[a(n-1)^2+a(n-1)],整理得[an+a(n-1)][a
因为2√S(n)=a(n)+12√S(n+1)=a(n+1)+1所以两式平方相减4(S(n+1)-S(n))=[a(n+1)+1]^2-[a(n)+1]^24·a(n+1)=[a(n+1)]^2+2·
让我来详细解答吧:(1)Sn²=an(Sn-1)Sn²=[sn-s(n-1)]*(sn-1)=Sn²-sn*sn(n-1)-sn+sn(n-1)sn-sn(n-1)=-s
Sn²=an(Sn-1)Sn²=[sn-s(n-1)]*(sn-1)=Sn²-sn*sn(n-1)-sn+sn(n-1)sn-sn(n-1)=-sn*sn(n-1)两边同
解题思路:方法:数列通项的求法:已知sn,求an。求和:错位相减法。解题过程:
an=n(2^n-1)an=n*2^n-na1=1*2^1-1a2=2*2^2-2a3=3*3^3-3.an=n*2^n-nSn=a1+a2+a3+.+an=1*2^1-1+2*2^2-2+3*3^3
Sn-S(n-1)=2An-2A(n-1)=An所以An=2A(n-1)An/2A(n-1)=2即An为等比为2的等比数列令n=1,S1=3+2A1=A1A1=-3所以An=-3*[2^(n-1)]