Question

｛an}前n项和为Sn，Sn=4/3an-1/3*2^（n+1）+2/3.(1)求an (2)若Tn=2^n/Sn,求证：T1+T2+……Tn<3/2

求详细过程。谢谢！！！

Answer 1

(1)解：
当n=1时，a1=S1=(4/3)a1-(1/3)*2^(1+1)+2/3=(4/3)a1-2/3，解得:a1=2；
当n>1时：
Sn=(4/3)an-(1/3)*2^(n+1)+2/3=(4/3)an-2*(1/3)*2^n+2/3
S(n-1)=(4/3)a(n-1)-(1/3)*2^n+2/3=(4/3)a(n-1)-1*(1/3)*2^n+2/3
an
=Sn-S(n-1)
=[(4/3)an-2*(1/3)*2^n+2/3]-[(4/3)a(n-1)-1*(1/3)*2^n+2/3]
=(4/3)an-(4/3)a(n-1)-(1/3)*2^n
∴(1/3)an=(4/3)a(n-1)+(1/3)*2^n
即 an=4*a(n-1)+2^n
4*a(n-1)=4^2*a(n-2)+4*2^(n-1)
……
4^(n-2)*a2=4^(n-1)*a1+4^(n-2)*2^2
上述式子相加，得：
an=4^(n-1)*a1+2^n+4*2^(n-1)+…+4^(n-2)*2^2
=2^(2n-2)*2+2^n+2^2*2^(n-1)+…+2^(2n-4)*2^2
=2^(2n-1)+2^n+2^(n+1)+…+2^(2n-2)
=2^(2n-1)+2^n[2^0+2^1+…+2^(n-2)]
=2^(2n-1)+2^n*2^0*[1-2^(n-1)]/(1-2)
=2^(2n-1)+2^n*[2^(n-1)-1]
=2^(2n-1)+2^(2n-1)-2^n
=2^1*2^(2n-1)-2^n
=2^(2n)-2^n
∵a1=2=2^2-2^1，符合上式
∴数列{an}的通项公式是an=2^(2n)-2^n.
(2)证明：
Sn=(2^2-2^1)+(2^4-2^2)+…+[2^(2n)-2^n]
=[2^2+2^4+…+2^(2n)]-(2^1+2^2+…+2^n)
=4[1-(2^2)^n]/(1-2^2)-2(1-2^n)/(1-2)
=(4/3)[(2^n)^2-1]-2(2^n-1)
=(4/3)*(2^n)^2-4/3-2*2^n+2
=(4/3)*(2^n)^2-2*2^n+2/3
则Tn=2^n/Sn=1/[(4/3)*(2^n)-2+2/(3*2^n)]=(3/2)*1/(2*2^n+1/2^n-3).
设f(n)=1/(2*2^n+1/2^n-3)
=(2^n)/[2*(2^n)^2+1-3*(2^n)]
=(2^n)/(2^n-1)(2*2^n-1)
=[(2*2^n-1)-(2^n-1)]/(2^n-1)(2*2^n-1)
=1/(2^n-1)-1/[2^(n+1)-1]
则Tn=(3/2)*f(n)=(3/2)*{1/(2^n-1)-1/[2^(n+1)-1]}.
T1+T2+T3+…+Tn=(3/2)*{(1-1/3)+(1/3-1/7)+(1/7-1/15)+…1/(2^n-1)-1/[2^(n+1)-1]}
=(3/2)*{1-1/[2^(n+1)-1]}
=3/2-(3/2)*{1/[2^(n+1)-1]}
<3/2

Answer 2

解：(1) Sn=4/3an-1/3*2^（n+1）+2/3，S(n+1)=4/3a(n+1)-1/3*2^（n+2）+2/3，两式相减得
a(n+1)=4an+2^(n+1)，两边同除以2^(n+1) 得 a(n+1)/ 2^(n+1)=2an/2^n+1
令bn= an/2^n，得 b(n+1)=2bn+1，两边同加1得b(n+1)+1=2(bn+1)，∴﹛bn+1﹜是首项为b1+1，公比为2的等比数列，由Sn=4/3an-1/3*2^（n+1）+2/3得
a1=4/3a1-1/3*2^(1+1)+2/3，即a1=2，b1=a1/2=1，∴bn+1=( b1+1)* 2^(n-1)= 2^n
∴bn= 2^n-1，an=2^n*bn=2^2n－2^n
(2) 把an=2^2n－2^n代入Sn=4/3an-1/3*2^（n+1）+2/3得
Sn=4/3*2^2n -2*2^n+2/3=2/3(2^n-1)( 2*2^n-1)，
∴Tn=2^n/Sn=3*2^n/2(2^n-1)( 2*2^n-1)=3/2[1/(2^n-1)－1/(2*2^n-1)]
故Tn=3/2[1－1/3+1/3－1/7＋1/7－……＋1/(2^n-1)－1/(2*2^n-1)]
=3/2[1－1/(2*2^n-1)] <3/2