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
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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
![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](/uploads/image/z/3490243-43-3.jpg?t=%26nbsp%3B%E6%AD%A3%E9%A1%B9%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%EF%BC%8Ca4%3D16%EF%BC%8C%E4%B8%94a2%2Ca3%E7%9A%84%E7%AD%89%E5%B7%AE%E4%B8%AD%E9%A1%B9%E4%B8%BAS2.%281%29%E6%B1%82%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC)
(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*
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*
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