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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,

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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
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,
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?