Question

在数列{an}中，若Sn=4/3an-1/3*2^(n+1)+2/3，证明：数列{an+2^n}是

在数列{an}中，若Sn=4/3an-1/3*2^(n+1)+2/3，证明：数列{an+2^n}是等比数列，并求通项an，设bn=2^n/Sn，且前n项和为Tn，证明：Tn＜3/2

Answer 1

解：
(1)
n=1时，a1=S1=(4/3)a1-(1/3)·2²+2/3
解得a1=2
n≥2时，an=Sn-S(n-1)=(4/3)an -(1/3)·2^(n+1) +2/3-[(4/3)a(n-1)-(1/3)·2ⁿ+2/3]
整理，得
an=4a(n-1)+2ⁿ
an+2ⁿ=4a(n-1)+2^(n+1)=4a(n-1)+4·2^(n-1)=4[a(n-1)+2^(n-1)]
(an+2ⁿ)/[a(n-1)+2^(n-1)]=4，为定值
a1+2=2+2=4，数列{an +2ⁿ}是以4为首项，4为公比的等比数列
an+2ⁿ=4ⁿ
an=4ⁿ-2ⁿ
数列{an}的通项公式为an=4ⁿ-2ⁿ
Sn=(4/3)an-(1/3)·2^(n+1) +2/3
=(4/3)·(4ⁿ-2ⁿ)-(2/3)·2ⁿ+2/3
=[4^(n+1) -3·2^(n+1) +2]/3
(2)
bn=2ⁿ/Sn
=3·2ⁿ/[4^(n+1)-3·2^(n+1)+2]
=3·2ⁿ/[4·(2ⁿ)²-6·2ⁿ+2]
=(3/2)·2ⁿ/[2·(2ⁿ)²-3·2ⁿ+1]
=(3/2)·2ⁿ/[(2ⁿ-1)(2·2ⁿ-1)]
=(3/2)[1/(2ⁿ -1) -1/(2^(n+1) -1)]
Tn=b1+b2+...+bn
=(3/2)[1/(2-1)-1/(2²-1)+1/(2²-1)-1/(2³-1)+...+1/(2ⁿ-1)-1/(2^(n+1)-1)]
=(3/2)[1- 1/(2^(n+1) -1)]
=3/2 -(3/2)/[2^(n+1) -1]
(3/2)/[2^(n+1) -1]>0 3/2-(3/2)/[2^(n+1)-1]<3/2
Tn<3/2