设数列｛an｝的前n项和为Sn=2n^2,｛bn｝为等比数列,且a1=b1,b2(a2-a1)=b1.
问题描述：
设数列｛an｝的前n项和为Sn=2n^2,｛bn｝为等比数列,且a1=b1,b2(a2-a1)=b1.
(1)求数列｛an｝和｛bn｝的通项公式；
（2）设Cn=an/bn,求数列｛Cn｝的前n项和Tn.
答
a1=S1=2Sn=2n^2Sn-1=2(n-1)^2=2n^2-4n+2an=Sn-Sn-1=2n^2-2n^2+4n-2=4n-2n=1代入4-2=2=a1,同样满足.数列{an}通项公式为an=4n-2b1=a1=2a2=4×2-2=6b2(a2-a1)=b1b2(6-2)=2b2=1/2b2/b1=(1/2)/2=1/4数列{bn}是以2为首项,1...