设sn是正项数列an的前n项和,sn=1 3an2+1 2an
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/26 03:01:46
因为:(5n-8)Sn+1-(5n+2)Sn=-20n-8...(1)所以:(5(n+1)-8)Sn+2-(5(n+1)+2)Sn+1=-20(n+1)-8即:(5n-3)Sn+2-(5n+7)Sn+
(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+
(1)由已知有:2a1=4096得a1=2048,又an+sn=4096,an+1+Sn+1=4096,两式相减得an+1=an/2,所以an是以1/2为公比的等比数列,故an=2048*(1/2)^
∵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
∵数列{an}的前n项和为Sn=43an-13×2n+1+23(n=1,2,3…),∴当n=1时,a1=S1=43a1−13×22+23,解得a1=2.当n≥2时,an=Sn-Sn-1=43an-13
题目打完啊!,浪费资源
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
∵an=Sn-Sn-1(n≥2),Sn=n2∴a8=S8-S7=64-49=15故答案为15
解题思路:分析与答案如下,如有疑问请添加讨论,谢谢!点击可放大解题过程:最终答案:略
2^(n+1)-2^n=2*2^n-2^n=2^nb*an-2^n=(b-1)Sn,b*a(n+1)-2^(n+1)=(b-1)S(n+1)两式相减(左-左=右-右):[b*a(n+1)-2^(n+1
(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
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
Sn+1=4an+2Sn=4a(n-1)+2相减得Sn+1-Sn=4an+2-4a(n-1)-2an+1=4an-4a(n-1)an+1-2an=2(an-2an-1)bn=2bn-1(2)求数列{a
/>n≥2时,an=Sn/n+2(n-1)Sn=nan-2n(n-1)S(n-1)=(n-1)an-2(n-1)(n-2)Sn-S(n-1)=an=nan-2n(n-1)-(n-1)an+2(n-1)
(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)∵an+Sn=4096,∴a1+S1=4096,a1=2048.当n≥2时,an=Sn-Sn-1=(4096-an)-(4096-an-1)=an-1-an∴anan−1=12an=2048(1
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(
An+1=1/3Sn3An+1=Sn(1)3An=Sn-1(2)(1)-(2)得3An+1=4An(n大于等于2),所以An是以A2为首项q=4/3的等比数列A2=1/3A1,所以A2等于1/3An=
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: