Question

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.

Answer 1

(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