Question

高中数列题 求解！

正项等比数列{an}的前n项和为Sn，a4=16，且a2,a3的等差中项为S2. (1)求数列{an}的通项公式；(2)设bn=n／a2n-1，求数列{bn}的前n项和Tn

Answer 1

(1)2S2=a2+a3,2a1+2a2=a2+a3,化简得2a1+a2=a3,即2a1+a1*q=a1*q^2,因为an>0,a1>0,q>0,
2+q=q^2,q=2或-1，q=2又因为a4=a1*q^3=16,a1=2,所以an=2^n,n∈N*
(2)bn=n/2^2n-1=2n/2^2n
Tn=2/2^2+2*2/2^4+2*3/2^6+-------+2n/2^2n
Tn/2^2= 2/2^4+2*2/2^6+2*3/2^8+--------+2(n-1)/2^2n+2n/2^(2n+2)
两式相减，得3/4Tn=2/2^2+2/2^4+2/2^6+-------2/2^2n-2n/2^(2n+2)
3/4Tn=1/2+1/2^3+1/2^5+------+1/2^(2n-1)-n/2^(2n+1)=1/2*(1-1/2^n)/1-1/4-n/2^(2n+1)
Tn=8/3-8/3*2^n-n/2^(2n+1),n∈N*

Answer 2

a2+a3=2S2=S3-a1