设数列an的前n项和为sn,对任意n∈N*满足2Sn=an(an 1)
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an+Sn=4096=2^12an-1+sn-1=2^12an-an-1+(sn-sn-1)=02an=an-1an/(an-1)=1/2q=1/2a1=s1=2^11an=2^11(1/2)^(n-
an+Sn=4096a1+s1=4096a1=2048=2^11Sn=4096-anS(n-1)=4096-a(n-1)两式相减得an=a(n-1)-anan=(1/2)a(n-1){an}是公比为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+
an+Sn=4096,a(n+1)+S(n+1)=4096,相减得2a(n+1)=an,{an}是等比数列an=2^(12-n),得logan=(12-n)log2是等差数列Tn=n[11+12-n]
(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)^
an+Sn=4096a(n-1)+S(n-1)=4096两式相减==>>an=a(n-1)/2(又a1+s1=4096=>a1=2048)==>>an=2^(12-n)==>>bn=(log2底an)
an+Sn=4096a(n+1)+S(n+1)=4096相减a(n+1)-an+a(n+1)=0a(n+1)/an=1/2所以是等比,q=1/2a1=S1所以2a1=4096a1=2048=2^11所
an=Sn-Sn-1=n(a1+an)/2-(n-1)(a1+an-1)/22an=na1+nan-na1-nan-1+a1+an-1(n-2)an=(n-1)*(an-1)-a1(1)同理(n-1)
(1)an+Sn=na(n+1)+S(n+1)=n+1两式相减2a(n+1)-an=1,即2(a(n+1)-1)=an-1,2b(n+1)=bn而a1+a1=1,a1=1/2,b1=-1/2,{bn}
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
(1)(an+2)/2=根号下2Sn所以8Sn=(an+2)^2n=1,S1=a1.8a1=(a1+2)^2,得a1=2n=2,8S2=(a2+2)^2,8(a1+a2)=(a2+2)^2,得a2=6
题目中(am-an)/(m+n)是错的,应改为(am-an)/(m-n).必要性:an是公差为d的等差数列,则am=a1+(m-1)d,an=a1+(n-1)d,2S(m+n)=2(m+n)a1+(m
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
2Sn=an(an+1),2Sn=a(n-1)【a(n-1)+1】,an=Sn-S(n-1)得2an=an^2(平方)-a(n-1)^2+an-a(n-1).移项,平方的用平方差,因为an≠0,所以两
/>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
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: