a1=-1,an_ 1=Sn*Sn 1

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)

推倒正确,你的问题是?

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