在等比数列{an}中,Sn=2^n-1,求则数列{an²}

an,Sn,Sn-1/2成等比数列an(Sn-1/2)=Sn^2a2(S2-1/2)=S2^2a2(a2+1/2)=(a2+1)^2a2=-2/3a3(S3-1/2)=S3^2a3(a3-1/6)=(

因为an,Sn,Sn-1/2成等比数列Sn（平方）=an*（Sn-1/2）由an=Sn-S（n-1）Sn（平方）=（Sn-S（n-1））*（Sn-1/2）化简得S（n-1）*Sn=S（n-1）/2-S

a1=S1=2^1-1=1a2=S2-S1=2^2-1-1=2公比q=a2/a1=2/1=2an是等比数列——首项是1,公比是2an^2也是等比数列——首项是a1^2=1,公比是q^2=4a1^2+a

设{an}的公比为q,则a2=2q,a3=2q^2则(a2+1)^2=(a1+1)(a3+1)即(2q+1)^2=3(2q^2+1)解得q=1所以{an}为常数数列Sn=na1=2n

因数列{an}为等比，则an=2qn-1，因数列{an+1}也是等比数列，则（an+1+1）2=（an+1）（an+2+1）∴an+12+2an+1=anan+2+an+an+2∴an+an+2=2a

Sn,S2n-Sn,S3n-S2n成等比数列48,12,3S3n-S2n=3S3n=3+S2n=63

Sn=a1(1-q^n)/(1-q)S1=a1S2=a1(1+q)S3=a1(1+q+q^2)S2+2=a1(1+q)+2S3+2=a1(1+q+q^2)+2[a1(1+q+q^2)+2]*[a1+2

k=-3由等比前n项和公式得Sn=a1(1-q^n)/(1-q)=-[a1/(1-q)]×q^n+a1/(1-q)前面系数与后面的常数相反,∴k=-3

a5=4,a7=6,a5=a1*q^4=4a7=a1*q^6=6q^2=3/21.q=根号6/22.q=-根号6/2a1=9a1=9an=a1*q^(n-1)an=9*(根号6/2)^(n-1)an=

因为a1＝2a4＝−54，所以q3=-27，所以q=-3，所以an=2×（-3）n-1Sn＝2[1−(−3)n]1−(−3)＝1−(−3)n2

an,Sn,Sn-1/2成等比数列an(Sn-1/2)=Sn^2a2(S2-1/2)=S2^2a2(a2+1/2)=(a2+1)^2a2=-2/3a3(S3-1/2)=S3^2a3(a3-1/6)=(

因数列{an}为等比，则an=3qn-1，因数列{an+1}也是等比数列，则（an+1+1）2=（an+1）（an+2+1）∴an+12+2an+1=anan+2+an+an+2∴an+an+2=2a

an=a1*q^(n-1)96=a1*2^(n-1)192=a1*2^nSn=(a1-a1q^n)/(1-q)189=(a1-192)/(1-2)189=-a1+192a1=3192=3*2^n64=

设等比数列{an}的公比为q，则可得an=2•qn-1，故an+1=2•qn-1+1，可得a1+1=3，a2+1=2q+1，a3+1=2q2+1，由于数列{an+1}也是等比数列，故（2q+1）2=3

因为6Sn=(an+1)(an+2)（1）所以6Sn-1=(an-1+1)(an-1+2)（2）（1）-（2）则an-an-1=3所以an是等差数列因为6Sn=(an+1)(an+2)可知S1＝a1=

a1=S1=k-4a1+a2=S2=k-8所以,a2=(k-8)-(k-4)=-4a1+a2+a3=S3=K-16所以,a3=(k-16)-(k-8)=-8所以,(-8)*(k-4)=(-4)^2=1

Sn=n-2an,Sn-1=（n-1）-2an-1（n大于1）做差an=1-2an-2an-13an-3=2an-1-2（an-1）/[a（n-1）-1]=2/3是常数,经检验,a1=1/3,a2=5

(a2+1)²=(a1+1)(a3+1)a1=2,设an公比q(2q+1)²=3(2q²+1)4q²+4q+1=6q²+32q²-4q+2=

设公比为q,a2²=a1*a3(a2+1)²=(a1+1)(a3+1)因为a1=2所以a2²=2a3(a2+1)²=3(a3+1)解得a2=2a3=2所以sn=

已知Sn=2An-1取n=1得:S1=2A1-1又因为S1=A1,解上述方程可得:A1=1Sn=2An-1S(n-1)=2A(n-1)-1注:"n-1"为下标上下两式相减得:Sn-S(n-1)=2An