sn不为零,an=sn^2 sn-1,求an
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1.Sn=-2an+3有S(n-1)=-2a(n-1)+3则an=Sn-S(n-1)=-2an+2a(n-1)=>an=a(n-1)*2/3所以,{an}为共比数列,q=2/32.Sn=-2an+3有
证明:(1)证:因为Sn=4an-p(n∈N*),则Sn-1=4an-1-p(n∈N*,n≥2),所以当n≥2时,an=Sn-Sn-1=4an-4an-1,整理得an=43an−1.(5分)由Sn=4
1.因为等差数列AN的公差d不等于0,a1=2,s9=36,所以36=9*2+1/2*9*8d所以d=1/2所以a3=3,a9=6,由a3,a9,am成等比数列则a9的平方=a3*am,的am=12又
答案为ASn=((a1+an)/2)*nan=a1+(n-1)d根据上式得出:Sn=(2a1+(n-1)d)*n/2=a1*n+n方*d/2-n*d/2limSn/n方=lim(2a1*n+n方*d-
(1)∵{An}为等比数列,则有An+1=An·q,又∵Sn+1,Sn,Sn+2成等差数列,∴Sn+1+Sn+2=2Sn∴Sn+An+Sn+An+An·q=2Sn∴可得2+q=0所以q=-2(2)这里
因为Sn+1,Sn,Sn+2成等差数列S(n+1)+S(n+2)=2*S(n)(q^(n+1)-1)*a1/(q-1)+(q^(n+2)-1)*a1/(q-1)=2*(q^(n)-1)*a1/(q-1
∵Sn-Sn-1=√Sn+√Sn-1∴(√Sn)²-(√Sn-1)²=√Sn+√Sn-1(√Sn-√Sn-1)(√Sn+√Sn-1)=√Sn+√Sn-1∴√Sn-√Sn-1=1(n
当n=1时、有2s1+1=3a1,即有a1=1,因为2Sn+1=3an,所以2Sn+1+1=3an+1.后式减去前式,得2an+1=3an+1-3an.即有an+1=3an,为等比数列,且公比为3,所
an+2Sn*Sn-1=0其中an=Sn-Sn-1代入上式:Sn-Sn-1+2Sn*Sn-1=0a1=1/2,故Sn和Sn-1≠0,上式两边同除以Sn*Sn-1得:1/Sn-1-1/Sn+2=0即:1
由Sn=Sn-1/2Sn-1+1,两边同时取倒数可得1/Sn=(2Sn-1+1)/Sn-11/Sn=2+1/Sn-1即1/Sn-1/Sn-1=2故{1/Sn}是首项为1/2,公差为2的等差数列1/Sn
第一个搞定我就不罗嗦了即1/Sn-1/Sn-1=2所以有1/Sn-1/Sn-1=21/Sn-1-1/Sn-2=21/Sn-2-1/Sn-3=2…………1/S2-1/S1=2叠加得1/Sn-1/S1=2
an=Sn·Sn-11=an/(Sn·Sn-1)=1/Sn-1-1/Sn1=1/Sn-2-1/Sn-1……………………1=1/S1-1/S2n-1式相加有n-1=1/S1-1/SnS1=a1所以1/S
Sn=a1(1-q^n)/(1-q)Sn+1=a1[1-q^(n+1)]/(1-q)Sn+2=a1[1-q^(n+2)]/(1-q)2Sn+2=Sn+Sn+1a1[1-q^(n+1)]/(1-q)+a
我会我会Sn+1=Sn-2nSn+1Sn两边同除以Sn+1*Sn得1/Sn+1-1/Sn=2n以此类推1/Sn-1/Sn-1=2(n-1)1/Sn-1-1/Sn-2=2(n-2)...1/S2-1/S
Sn^2-n^2×Sn-(n^2+1)=0(Sn+1)[Sn-(n^2+1)]=0数列各项为非零实数,S1≠0,且Sn不恒为0,因此只有Sn=n^2+1n=1时,a1=S1=1+1=2n≥2时,an=
由题意,S(n)-S(n-1)=2a(n+1)-2a(n),即a(n)=2a(n+1)-2a(n),于是a(n+1)=a(n)*3/2,即a(n)是公比是q=3/2的等比数列,且首项是a(1)=1,所
(1)由sn=sn-12sn-1+1(n≥2),a1=2,两边取倒数得1Sn=1Sn-1+2,即1Sn-1Sn-1=2.∴{1sn}是首项为1S1=1a1=12,2为公差的等差数列;(2)由(1)可得
因为Sn+Sn-1=3an所以Sn-1+Sn-1+an=3an2Sn-1=2anSn-1=an因为Sn=an+1所以Sn-Sn-1=an+1-anan=an+1-an2an=an+1an+1/an=2
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
解题思路:考查数列的通项,考查等差数列的证明,考查数列的求和,考查存在性问题的探究,考查分离参数法的运用解题过程: