已知an收敛:,其部分和为sn
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设∑an收敛到SS,n->∞∴1/Sn->1/S≠0,∴∑(1/Sn)发散
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
4Sn=(an+1)^24Sn-1=(an-1+1)^2n-1为下标则4an=4Sn-4Sn-1=(an+1)^2-(an-1+1)^2化简得(an-1)^2=(an-1+1)^2则an-1=正负(a
(1)2Sn=an^2+an2Sn-1=a(n-1)^2+a(n-1)2an=2Sn-2Sn-1=an^2-a(n-1)^2+an-a(n-1)an^2-a(n-1)^2=an+a(n-1)[an+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)
(1)(an+2)/2=根号下2Sn所以8Sn=(an+2)^2n=1,S1=a1.8a1=(a1+2)^2,得a1=2n=2,8S2=(a2+2)^2,8(a1+a2)=(a2+2)^2,得a2=6
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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
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
∵数列{an},其前n项和为sn,且sn=n2+n,∴当n≥2时,sn-1=(n-1)2+(n-1),∴an=sn-sn-1=(n2+n)-[(n-1)2+(n-1)]=2n;当n=1时,a1=s1=
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
已知等比数列an,首项为81,数列bn满足bn=log3an,其前n项和sn(1)证明:bn-b(n-1)=log(3)an-log(3)an-1=log(3)an/a(n-1)=log(3)q∵b1