题目

设数列an的前n项和Sn=4/3an-1/3*2n+1+2/3,n=1,2,3…….
（1）求首项a1与通项an；
（2）设Tn=2n/Sn,n=1,2,3,……,证明∑Ti

提问时间：2020-05-23

答案

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]}.
∴n
∑ Ti=T1+T2+T3+…+Tn
i=1
=(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]}