已知单调递增的等比数列｛an｝满足a1+a2+a3=14,且a2+1是a1、a3的等差中项.

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(1)
an=a1.q^(n-1) ; q>1
a1+a2+a3=14
a1(1+q+q^2)=14 (1)
(a2+1)是a1、a3的等差中项
a1+a3=2(a2+1)
a1(1+q^2) =2(a1q+1)
a1(1-2q+q^2) = 2 (2)
(1)/(2)
1+q+q^2=7(1-2q+q^2)
2q^2-5q+2=0
(2q-1)(q-2)=0
q=2
from (1)
a1(1+2+4)=14
a1=2
an =2^n
(2)
let
S = 1.2^1+2.2^2+...+n.2^n (1)
2S = 1.2^2+2.2^3+...+n.2^(n+1) (2)
(2)-(1)
S =n.2^(n+1) -(2+2^2+...+2^n)
=n.2^(n+1) -2(2^n-1)
bn=an.logan
=n.2^n
Sn=b1+b2+...+bn
=S
=n.2^(n+1) -2(2^n-1)
=2+ (2n-2).2^n
(3)
S(n+1)-2 ≤ 8n^3.λ
To find min λ
Solution:
S(n+1)-2
= 4n.2^n
S(n+1)-2 ≤ 8n^3.λ
4n.2^n ≤ 8n^3.λ
λ ≥ 2^(n-1) / n^2
min λ at n=3
min λ = 4/9
再问：最后一步是怎么的出来的啊
λ ≥ 2^(n-1) / n^2
min λ at n=3
min λ = 4/9
再答： f(x) = 2^(x-1)/ x^2
f'(x) = 2^(x-1) . [ ln2/x^2 - 2/x^3 ] =0
xln2 - 2 =0
x = 2/ln2 =2.88 (min)

f(2) = 2/4 = 1/2
f(3) = 4/9 < f(2)

min 2^(n-1)/n^2 at n=3

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