已知正项数列an中,其前n项和为Sn,且an=2√Sn-1
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(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)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
由题意得,Sn=[(an+1)/2]^2①则S(n+1)=[(a(n+1)+1)/2]^2②②-①得(结合a(n+1)=S(n+1)-Sn)a(n+1)=[(a(n+1)+1)/2]^2-[(an+1
sn=3*3^1+5*3^2+.+(2n+1)*3^n①3sn=3*3^2+5*3^3+.+(2n-1)*3^n+(2n+1)*3^(n+1)②①-②-2Sn=Sn-3Sn=-2n*3^(n+1),因
(1)等比数列{an},首项为81设an=a1*q^(n-1)=81*q^(n-1)数列{bn}满足bn=log3为底an∴bn=log3为底[81*q^(n-1)]=log3为底81+log3为底q
Sn=(an+1)^2/4=(an^2+2an+1)/4Sn-1=[a(n-1)+1]^2=[(a(n-1)^2+2a(n-1)+1]/4Sn-Sn-1=an=[an^2+2an-a(n-1)^2-2
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)
∵2根号Sn=an+14Sn=(an+1)^2①4S(n-1)=[a(n-1)]^2②①-②,可得:4an=[an^2-a(n-1)^2]+2[an-a(n-1)]化简可得:2[a(n-1)+an]=
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
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*
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)-
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
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