a1=-1,an_ 1=Sn*Sn 1
来源:学生作业帮助网 编辑:作业帮 时间:2024/10/06 08:59:30
1、S(n+1)=2Sn+3n+1S(n+1)+3(n+1)+4=2Sn+6n+8=2(Sn+3n+4)[S(n+1)+3(n+1)+4]/(Sn+3n+4)=2,为定值.S1+3+4=a1+3+4=
S(n+1)=2Sn+3n+1则S(n+1)-Sn=Sn+3n+1即a(n+1)=Sn+3n+1所以Sn=a(n+1)-3n-1所以S(n-1)=an-3(n-1)-1用上式减下式:Sn-S(n-1)
1).an+1=3(a(n-1)+1)令bn=an+1故有bn=3b(n-1),且b1=a1+1=2bn=3^(n-1)*2故,an=bn-1=3^(n-1)*2-12)令(an+t)=1/2(a(n
(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-
∵Sn-S(n-1)=√Sn-√S(n-1)∴Sn≥0(n≥2)又S1=a1=1∴Sn≥0(n≥1)又Sn-S(n-1)=[√Sn+√S(n-1)]*[√Sn-√S(n-1)]=√Sn-√S(n-1)
哎呀,大家不要乱答啊,错了好多第一题Sn×SQR(S(n-1))-S(n-1)SQR(Sn)=2SQR(Sn×S(n-1)所以SQR(S(n-1)*S(n))*(SQR(S(n)-SQR(S(n-1)
因为2S(n+1)=2a1+Sn2S(n+1)=2+Sn令2【S(n+1)+x】=Sn+x解得x=-2所以【Sn-2】这个数列是以S1-2=-1为首项以1/2为公比的等比数列所以Sn-2=(-1)*2
1、Sn=(1-(-32)*(-2))/(1+2)=-212、Sn-qSn=a1-anq(an-Sn)q=a1-Snq=(a1-Sn)/(an-Sn)
由题意得:2S(n+1)=4Sn+a1,则2Sn=4S(n-1)+a1解得:a(n+1)=2an,则{an}为等比数列,公比q=2所以,an=a1q^(n-1)=2^n同样:2S(n+1)=4Sn+a
(1)由2an=Sn*S(n-1),an=Sn-S(n-1)则:2[Sn-S(n-1)]=Sn*S(n-1)2Sn-2S(n-1)=Sn*S(n-1)两边同时除以Sn*S(n-1)2/S(n-1)-2
√Sn-√S(n-1)=√2令bn=√Sn则bn是以√2位公差的等差数列bn=b1+(n-1)√2S1=a1=2所以b1=√S1=√2所以bn=√2+(n-1)√2=n*√2所以Sn=(bn)^2=2
Sn+S(n+1)=5(a(n+1))/3因为S(n+1)=SN+A(N+1)所以Sn+SN+A(N+1)=5a(n+1)/32SN=2a(n+1)/3SN=a(n+1)/3S(N-1)=AN/3SN
两边同时除以sn得1=1/2(sn-1)+1/sn设1/sn-a=-1/2(1/(sn-1)-a)解得a=2/3,又a1=2,所以1/s1-2/3=-1/6所以1/sn-2/3=(-1/6)(-1/2
证明:(1)当n=1时左边=S1=a1=1右边=(2^1-1)/[2^(1-1)]=1左边=右边所以不等式成立(2)假设当n=k时等式成立即Sk=(2^k-1)/[2^(k-1)]那么当n=k+1时因
1)s3/S1=1得s3=s1又a1=1所以a3=1得an=n-12)Sn=n^2/2Bn=2/n^23)Tn
由S(n+1)/S(n)=(4n+2)/(n+1),可得a(n+1)/S(n)=S(n+1)/S(n)-1=(3n+1)/(n+1),所以S(n)=(n+1)/(3n+1)*a(n+1)以n-1代替n
S(n+1)=2Sn+a1.(1)Sn=2S(n-1)+a1.(2)(1)-(2)得S(n+1)-Sn=2[Sn-S(n-1)]a(n+1)=2an∴an是q=2的等比数列an=a1X2^(n-1)S
当n≥2时,可以化为Sn-S(n-1)=-2Sn×S(n-1),两边同除以Sn×S(n-1),得1/Sn-1/S(n-1)=2所以{1/Sn}是以2为首项,2为公差的等差数列即1/Sn=2nSn=1/
A2*An-1=A1*An=128A1+An=66=>A1,An为方程xx-66x+128=0两根=>A1,An=2,64或64,2若A1=2,An=64=>q^(n-1)=32Sn=A1(1-q^n