设Sn是数列an的前n项和,Sn0,a1=1,an 1 2SnSn 1=0
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![设Sn是数列an的前n项和,Sn0,a1=1,an 1 2SnSn 1=0](/uploads/image/f/7250480-8-0.jpg?t=%E8%AE%BESn%E6%98%AF%E6%95%B0%E5%88%97an%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%2CSn0%2Ca1%3D1%2Can+1+2SnSn+1%3D0)
S(n+1)+Sn=2a(n+1),a(n+1)+2Sn=2a(n+1),2Sn=a(n+1),2S(n-1)=an相减:2an=a(n+1)-an,q=a(n+1)/an=3an=3*3^(n-1)
(Ⅰ)因为a1=S1,2a1=S1+2,所以a1=2,S1=2,由2an=Sn+2n知:2an+1=Sn+1+2n+1=an+1+Sn+2n+1,得an+1=sn+2n+1①,则a2=S1+22=2+
∵Sn=n-an,∴a(n+1)=S(n+1)-S(n)=(n+1)-a(n+1)-n+a(n)=1+a(n)-a(n+1);∴2a(n+1)=1+a(n);∴2a(n+1)-2=1+a(n)-2,即
(1)Sn^2=3n^2an+S(n-1)^2Sn^2-S(n-1)^2=3n^2an[Sn+S(n-1)][Sn-S(n-1)]=3n^2an[Sn+S(n-1)]*an=3n^2anan≠0所以S
S(n+1)=4an+2Sn=4a(n-1)+2,n≥2S(n+1)-Sn=a(n+1)=4an-4a(n-1)a(n+1)-2an=2(an-2a(n-1))令a(n+1)-2an=bnbn/b(n
n=an+1S(n+1)=2Sn+n+5.1Sn=2S(n-1)+n-1+5=2S(n-1)+n+4.2(1)-(2)得S(n+1)-Sn=2[Sn-S(n-1)]+1a(n+1)=2an+1a(n+
an=-Sn.S(n-1)Sn-S(n-1)=-Sn.S(n-1)1/Sn-1/S(n-1)=11/Sn-1/S1=n-11/Sn=nSn=1/n
1.sn=2an+ns(n-1)=2a(n-1)+n-1相减得an=2an-2a(n-1)+1整理得an-1=2[2a(n-1)-1]所以an-1是等比数列首项a1由a1=2a1+1得a1=-1所以a
证明:法一:令d=a2-a1.下面用数学归纳法证明an=a1+(n-1)d(n∈N).(1)当n=1时上述等式为恒等式a1=a1.当n=2时,a1+(2-1)d=a1+(a2-a1)=a2,等式成立.
(2)a(n+1)=s(n+1)-s(n)=[2a(n+1)-2^(n+1)]-[2a(n)-2^n]所以a(n+1)-2an=2^n,当然就是等比数列哦
证:第一种方法Sn+1=(n+1)[a1+a(n+1)]/2Sn=n(a1+an)/2Sn-1=(n-1)[a1+a(n-1)]/2a(n+1)=Sn+1-Sn=(n+1)[a1+a(n+1)]/2-
Sn^2=3n^2an+S(n-1)^2Sn^2-S(n-1)^2=3n^2an[Sn+S(n-1)][Sn-S(n-1)]=3n^2an[Sn+S(n-1)]*an=3n^2anan≠0所以Sn+S
设数列{an}的前n项和为Sn,Sn=a1(3n−1)2(对于所有n≥1),则a4=S4-S3=a1(81−1)2−a1(27−1)2=27a1,且a4=54,则a1=2故答案为2
证明:A(n+1)=Sn+3n+1,则An=S(n-1)+3n-2两式想减得A(n+1)-An=Sn+3n+1-(S(n-1)+3n-2)=An+3即A(n+1)+3=2(An+3)即(A(n+1)+
Sn=4An-3S(n-1)=4A(n-1)-3Sn-S(n-1)=An=4An-3-[4A(n-1)-3]=4an-3-4A(n-1)+3=4An-4A(n-1)3An=4A(n-1)An/A(n-
且S1=2,S<n1>-Sn=Sn2=bn这句话的意思没看明白!∵bn=Sn+2∴b(n+1)=S(n+1)+2b(n+1)-bn=S(n+1)-Sn=bn∴b(n+1)=2*bn则b(n+1)/bn
(1)当n=1时,T1=2S1-1因为T1=S1=a1,所以a1=2a1-1,求得a1=1(2)当n≥2时,Sn=Tn-Tn-1=2Sn-n2-[2Sn-1-(n-1)2]=2Sn-2Sn-1-2n+
1.A1=S1=2A1-2^1A1=2S2=A1+A2=2A2-2^2A2=6S3=S2+A3=2A3-2^3A3=16S4=S3+A4=2A4-2^4A4=402.Sn=2An-2^nS(n+1)=
因为(n,Snn)在y=3x-2的图象上,所以将(n,Snn)代入到函数y=3x-2中得到:Snn=3n−2,即{S}_{n}=n(3n-2),则an=Sn-Sn-1=n(3n-2)-(n-1)[3(
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: