设sn是数列an的前n项和若不等式an2 sn2.n2
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应该是“Sn=2an+Sn-1”吧?
(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+
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
∵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
设{A(n)}的通项公式为:A(n)=2+d(n-1){B(n)}的通项公式为:B(n)=2×q^(n-1)则{A(n)}的前n项和为:S(n)=[A(1)+A(n)]n/2=[4+d(n-1)]n/
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)
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
(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
(1)Sn=2an-3nn=1时,S1=a1,故有:a1=2a1-3,a1=3n>=2时,an=Sn-S(n-1)=2an-3n-[2a(n-1)-3(n-1)]=2an-2a(n-1)-3即:an=
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-
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)+
(1)∵Sn=2an-3n,对于任意的正整数都成立∴S(n-1)=2a(n-1)-3n-3两式相减,得a(n+1)=2a(n+1)-2an-3,即a(n+1)=2an+3∴a(n+1)+3=2(an+
S1/a1=1S2/a2-S1/a1=(2+d)/(1+d)-1=d/(1+d)S3/a3-S1/a1==(3+3d)/(1+2d)-1=(2+d)/(1+2d)2*d/(1+d)=(2+d)/(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-
因为a(n+1)=S(n+1)-Sn=(n+2/n)Sn,所以得到S(n+1)=(2n+2/n)Sn,即得到S(n+1)/(n+1)=2*Sn/n,就得到Sn/n是等比数列,且公比为2,首项S1/1=
设公比为q,因为a1=1,即:a(n)=q^(n-1)则:S(n)=(1-q^n)/(1-q)若{Sn}为等差数列,设公差为d则:S(n)=S(n-1)+d即:d=S(n)-S(n-1)=(1-q^n
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)=
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: