已知正项数列an的前n项和为sn,根号sn是1/4与(an+1)的等比中项.1,求证,an是等差数列.2,若b1=a1,

来源：学生作业帮编辑：拍题作业网作业帮分类：数学作业时间：2024/04/28 17:34:19

已知正项数列an的前n项和为sn,根号sn是1/4与(an+1)的等比中项.1,求证,an是等差数列.2,若b1=a1,且bn=2b(n-1)+3,求数列bn的通项公式.3,在2的条件下,若cn

a(n)>0.
s(n) = [a(n)+1]/4,a(1) = s(1) = [a(1)+1]/4,a(1) = 1/3.
s(n+1) = [a(n+1)+1]/4,
a(n+1) = s(n+1)-s(n) = [a(n+1)+1]/4 - [a(n)+1]/4 = a(n+1)/4 - a(n)/4,
a(n+1) = -a(n)/3,
{a(n)}是首项为a(1)=1/3,公比为-1/3的等比数列.
a(n) = (1/3)(-1/3)^(n-1) = -1/(-3)^n.
b(1)=a(1)=1/3.
b(n+1) = 2b(n) + 3,
b(n+1)+3 = 2b(n)+6 = 2[b(n)+3],
{b(n)+3}是首项为b(1)+3 = 3 + 1/3 = 10/3,公比为2的等比数列.
b(n) + 3 = (10/3)*2^(n-1),
b(n) = (5/3)2^n - 3
c(n) = a(n)/[b(n)+3] = -1/(-3)^n /[(5/3)2^n] = -1/[(-3)^n*(5/3)2^n] = -(3/5)/(-6)^n
= -3/5*(-1/6)*(-1/6)^(n-1)
= (1/10)(-1/6)^(n-1)
t(n) = (1/10)[1 - (-1/6)^n]/(1+1/6) = (3/35)[1 - (-1/6)^n]
再问：可是要证an为等差数列
再答：难道[s(n)]^(1/2)是1/4和[a(n)+1]的等差中项？
2[s(n)]^(1/2) = [1/4 + a(n)+1] = a(n) + 5/4,
4s(n) = [a(n) + 5/4]^2 = [a(n)]^2 + 5a(n)/2 + 25/16,
4a(1) = 4s(1) = [a(1)]^2 + 5a(1)/2 + 25/16,
0 = [a(1)]^2 - 3a(1)/2 + 25/16,
Delta = (3/2)^2 - 4*25/16 = 9/4 - 25/4 < 0.
a(1)无解啊。。这也不对啊。。
楼主。。[s(n)]^(1/2)应该还是1/4和[a(n)+1]的等比中项。。
这样的话，a(n)确实是等比数列，而绝不是等差数列。。。
是不是老师写错题了？

已知正项数列an的前n项和为sn,根号sn是1/4与(an+1)的等比中项.1,求证,an是等差数列.2,若b1=a1, 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}是等差数列已知各项均为正数的数列{an}的前n项和为Sn,根号Sn是1/4与(an+1)^2的等比中项. 已知正项数列an的前n项和为Sn,a1=1,（an-2）²=8Sn-1.证明an是等差数列. 已知正项数列{an}的前n项为Sn,√Sn 是 1/4 与（an + 1）^2 的等比中项. 已知公差不为0的等差数列{An}的首项A1=1,前n项和为Sn,若数列{Sn/An}是等差数列,求An? 已知数列an的前n项和为Sn,且an+2Sn*Sn-1=0,a1=1/2,求证1/SN是等差数列,求数列SN的的通项公式已知数列{an}的前n项和为Sn,数列{Sn+1}是公比为2的等比数列,a2是a1和a3的等比中项已知数列An的前n项和为Sn.且满足an+2Sn*Sn-1=0=2>,a1=1/2,求证1/Sn是等差数列,求通项an的设数列an的前n项和为Sn,a1=1,an=(Sn/n)+2(n-1)(n∈N*) 求证:数列an为等差数列, 设数列an的前n项和为Sn,已知a1=1,Sn+1=4an+2 (1)设bn=an+1-2an,证明数列{bn}是等比数