Sn为an的前n项和,若不等式n平方an的平方 8Sn²≥
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/13 17:20:58
应该是“Sn=2an+Sn-1”吧?
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-
设等比数列{an}的公比为q,前n项和为Sn,且Sn+1,Sn,Sn+2成等差数列,则2Sn=Sn+1+Sn+2.若q=1,则Sn=na1,式子显然不成立.若q≠1,则有2a1(1−qn)1−q=a1
设:等差数列{an}的公差为d,通项为an=a1+(n-1)d,则:sn=a1+a2+...+an=na1+n(n-1)d/2lim(n->∞)(n*an)/Sn=lim(n->∞)[n*(a1+(n
∵等差数列{an}前m项和为Sm,若Sm:Sn=m^2:n^2∴m(a1+am)/n(a1+an)=m^2/n^2∴m[2a1+(n-1)d]=n[a1+(m-1)d]∴2(m-n)a1=(m-n)d
n=1时,a1=1+3a1.即a1=-1/2.n>1时,an=Sn-Sn-1=1+3an-(1+3a(n-1))=3an-3a(n-1),即an=3/2a(n-1),即an=-1/2*(3/2)^(n
Sn=n-5an-85(1)S(n+1)=n+1-5a(n+1)-85(2)(2)-(1)整理得6a(n+1)=1+5an即a(n+1)-1=(5/6)(an-1)又由S1=a1=1-5a1-85得a
∵an=1n(n+1)=1n−1n+1,∴S5=a1+a2+a3+a4+a5=1−12+12−13+13−14+14−15+15−16=56,故选B
A:1Sn=n(a1+an)/2,an^2+Sn^2/n^2=an^2+[(a1+an)/2]^2=[5an^2+2a1*an+a1^2]/4=(5/4)[an^2+(2/5)a1*an+a1^2/2
(1)证明:∵Sn=n-5an-85,n∈N*(1)∴Sn+1=(n+1)-5an+1-85(2),由(2)-(1)可得:an+1=1-5(an+1-an),即:an+1-1=56(an-1),从而{
(1)当n=1时,a1=S1=13(a1−1),得a1=−12;当n=2时,S2=a1+a2=13(a2−1),得a2=14,同理可得a3=−18.(2)当n≥2时,an=Sn−Sn−1=13(an−
对3a(n+1)+an=4变形得:3[a(n+1)-1]=-(an-1)a(n+1)/an=-1/3an=8*(-1/3)^(n-1)+1Sn=8{1+(-1/3)+(-1/3)^2+……+(-1/3
把Sn换成首项末项和项数的表达式,然后把末项看做变量,求最小值,可以得到m最大值为1/5(最小值为负无穷.)
Sn=a1+a2+a3+.+anS2n=a1+a2+a3+.+an+a(n+1)+.+a2ns2n-Sn=a(n+1)+a(n+2)+.+a2n=[1/2+1/3+1/4+.+1/(n+1)+1/(n
解题思路:方法:数列通项的求法:已知sn,求an。求和:错位相减法。解题过程:
n=1时an=s1=3n≥2时an=Sn-Sn-1=3^n-3^(n-1)=2*3^(n-1)
Sn=3an+2n可得S(n-1)=3a(n-1)+2n-2an=Sn-S(n-1)=3an+2n-3a(n-1)-2n+2即:an=3an-3a(n-1)+23a(n-1)=2an+2配项可得: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:
an+Sn=41a(n+1)+S(n+1)=2a(n+1)+Sn=422-1得2a(n+1)-an=0a(n+1)=1/2anan+Sn=4an≠0a(n+1)/an=1/2数列{an}是等比数列