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.

Question

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.

Answer 1

a1=S1=2
Sn=2n^2
Sn-1=2(n-1)^2=2n^2-4n+2
an=Sn-Sn-1=2n^2-2n^2+4n-2=4n-2
n=1代入4-2=2=a1,同样满足.
数列{an}通项公式为an=4n-2
b1=a1=2
a2=4×2-2=6
b2(a2-a1)=b1
b2(6-2)=2
b2=1/2
b2/b1=(1/2)/2=1/4
数列{bn}是以2为首项,1/4为公比的等比数列.
bn=2(1/4)^(n-1)=8/4^n
数列{bn}的通项公式为bn=8/4^n
cn=an/bn=(4n-2)/[8/4^n]=(2n-1)4^n/4=2n4^(n-1)-4^(n-1)
Tn=2[1×4^0+2×4^1+3×4^2+...+n×4^(n-1)]-(4^n-1)/(4-1)
令Mn=1×4^0+2×4^1+3×4^2+...+n×4^(n-1)
则4Mn=4^1+2×4^2+3×4^3+...+(n-1)×4^(n-1)+n×4^n
Mn-4Mn=-3Mn=4^0+4^1+4^2+...+4^(n-1)-n×4^n=(4^n-1)/(4-1)-n4^n
Mn=n4^n/3-(4^n-1)/9
Tn=2n4^n/3-2(4^n-1)/9-(4^n-1)/3
=[6n4^n-2×4^n+2-3×4^n+3]/9
=[(6n-2-3)4^n+5]/9
=[(6n-5)4^n+5]/9