若等差数列{an}的前n项和为Sn,且满足SnS2n为常数,则称该数列为S数列.
来源:学生作业帮 编辑:大师作文网作业帮 分类:数学作业 时间:2024/09/21 03:35:57
若等差数列{an}的前n项和为Sn,且满足
S
(Ⅰ)由an=4n-2,得a1=2,d=4,
Sn S2n= 2n+ 1 2n(n−1)4 2n•2+ 1 2•2n(2n−1)4= 1 4, 所以它为S数列; (Ⅱ)设等差数列{an},公差为d,则 Sn S2n= a1n+ 1 2n(n−1)d 2a1n+ 1 2•2n(2n−1)d=k(常数), ∴2a1n+n2d-nd=4a1kn+4n2dk-2nkd,化简得d(4k-1)n+(2k-1)(2a1-d)=0①, 由于①对任意正整数n均成立, 则 d(4k−1)=0 (2k−1)(2a1−d)=0解得: d=2a1≠0 k= 1 4., 故存在符合条件的等差数列,其通项公式为:an=(2n-1)a1,其中a1≠0.
如果数列an满足a{n+1}=pan+q(p,q为常数),则称an为"H数列".已知数列an的前n项和为Sn,若Sn=2
若等差数列{an}的首项为a1,公差为d,前n项的和为Sn,则数列(Sn/n)为等差数列,且通项
已知数列{An}的各项均为正数,前n项和为Sn,且满足2Sn=An²+n-4 1.求证{An}为等差数列
等差数列{An}的前n项和为Sn,且满足A3*A4=117,A2+A5=22 求通项An ;若数列{Bn}是等差数列,且
数列{an}满足a1=1,设该数列的前n项和为Sn,且Sn,Sn+1,2a1成等差数列.用数学归纳法证明:Sn=(2n-
已知数列{an}的前n项和为Sn,且满足an+2Sn*Sn-1=0,a1=1/2.求证:{1/Sn}是等差数列
已知数列{an}的前n项和为Sn,且满足Sn=Sn-1/2Sn-1 +1,a1=2,求证{1/Sn}是等差数列
各项均为正数的数列{an}的前n项和为S,且sn=1\8(an+2)².求证数列{an}是等差数列
求证等差数列!已知数列an的各项均为正数,前n项和为Sn,且满足2Sn=a∧2n+n-4
已知数列{an}得前n项和为sn=an^2+bn(a,b为常数且a不等于0)求证数列{an}是等差数列
已知数列{an}的前n项和为Sn,首项为a1,且1,an,Sn等差数列
设数列{an}的前n项和为Sn,满足2Sn=an+1-2^n+1+1,且a1,a2+5.a3成等差数列
|