Question

数列an的前n项和Sn且满足Sn=4/3an-1/3*2^n+1+2/3

,1)求数列首项a1与通项Sn(2)设数列bn=2^n/Sn前项和Tn求证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定值
a1+2=2+2=4数列{an
+2ⁿ}4首项4公比等比数列
an+2ⁿ=4ⁿ
an=4ⁿ-2ⁿ
数列{an}通项公式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)]
/主要拆项变形面简单了
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