已知数列an中,sn是其前n项和,并且sn 1=4an 2
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由Sn^2=an(Sn-1/2),两边同时除以Sn,拆开括号,得到Sn=an-an/2Sn,移项,an-Sn=an/2Sn,两边同时除以an,乘以2,得到2(an-Sn)/an=1/Sn,那么1/(S
(1)由[S(n)]^2=a(n)[S(n)-1/2]以及a(n)=S(n)-S(n-1),n≥2得[S(n)]^2=[S(n)-S(n-1)][S(n)-1/2],n≥2整理得2S(n)S(n-1)
(Sn)²=[Sn-S(n-1)](Sn-1/2)(Sn)²=(Sn)²-Sn/2-SnS(n-1)+S(n-1)/2Sn+2SnS(n-1)-S(n-1)=0S(n-1
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
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: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)
根据题意:首项为a1,公差为d3a8=5a13因为:a8=a1+7da13=a1+12d所以:3(a1+7d)=5(a1+12d)3a1+21d=5a1+60da1=-19.5d即:a20=a1+19
a3=a1+2d=6S3=a1+a2+a3=3a1+3d=12解得a1=2,d=2,故an=2n所以Sn=n(n+1)所以1/S1+1/S2+……+1/Sn=1/(1*2)+1/(2*3)+1/(3*
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
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(
题目是不是错了?经化简可得2Sn/Sn-1=1-(Sn-1/Sn),发现Sn/Sn-1无解
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·
哎,看你着急的样子,我就替你解了此因果S(n+1)=4an+2Sn=4a(n-1)+2相减得:a(n+1)=4an-4a(n-1)移向得a(n+1)-2an=2(an-2a(n-1)){a(n+1)-
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
因为Sn=3n^2+5nS(n-1)=3(n-1)^2+5(n-1)两式相减所以an=6n-3+5=6n+2所以an=8+6(n-1),所以an是以8为第一项,公差为6的等差数列.
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