在数列an中 其前n项和为sn 满足2sn=n-n方
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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)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
(1)an=sn-s(n-1)就有sn-s(n-1)+2sn*s(n-1)=0两边同除以sn*s(n-1)得1/sn-1/s(n-1)=2{1/sn}是等差数列1/sn=1/s1+(n-1)d=2n-
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.
S[1]=a[1]=1/2(a[1]+1/a[1]),于是:a[1]=1=√1-√0S[2]=a[2]+1=1/2(a[2]+1/a[2]),于是:a[2]=√2-1,S[2]=√2S[3]=a[3]
当公比为1时,Sn=n,数列{Sn+12}为数列{n+12}为公差为1的等差数列,不满足题意;当公比不为1时,Sn=1−qn1−q,∴Sn+12=1−qn1−q+12,Sn+1+12=1−qn+11−
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)
/>an=sn-s(n-1)=an+n^2-n-[a(n-1)+(n-1)^2-(n-1)]=an-a(n-1)+2n-2a(n-1)=2(n-1)an=2n所以,数列an为公差为2的等差数列.
1、an=Sn-S(n-1)所以2Sn-S(n-1)=20482Sn=S(n-1)+20482Sn-4096=S(n-1)+2048-40962(Sn-2048)=S(n-1)-2048(Sn-204
sn=2^n-11)n=1a1=2^1-1=12)n>=2sn-1=2^(n-1)-1an=2^n-2^(n-1)=2^(n-1)(n>=2)验证a1也满足,即an=2^(n-1)an^2=4^(n-
1.证:Sn=(3an-n)/2Sn-1=[3a(n-1)-(n-1)]/2an=Sn-Sn-1=[3an-3a(n-1)-1]/2an=3a(n-1)+1an+1/2=3a(n-1)+3/2=3[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·
若S10>0,则S10=(a1+a10)*10/2>0则2a1+9d>0.则d>-2a1/9同理S11
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
1.数列的第n项:a(n)=S(n)-S(n-1)=2a(n)-2a(n-1)移项得a(n)=2*a(n-1)所以n≥2时数列{a(n)}为公比q=2的等比数列;a(2)=S(2)-S(1)=2a(2
由题意得(an+1)/2=√(Sn×1)Sn=[(an+1)/2]²n=1时,S1=a1=[(a1+1)/2]²,整理,得(a1-1)²=0a1=1n≥2时,Sn=[(a
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