已知{an}是单调递增的等差数列,首项a1=3,前n项和为Sn,数列{bn}是等比数列,首项b1=1,且a2b2=12,S3+b2=20．（Ⅰ）求{an}和{bn}的通项公式．（Ⅱ）令Cn=Sncos（anπ）（n∈N+）,求{cn}的前n项和Tn．

Question

已知{an}是单调递增的等差数列,首项a1=3,前n项和为Sn,数列{bn}是等比数列,首项b1=1,且a2b2=12,S3+b2=20．（Ⅰ）求{an}和{bn}的通项公式．
（Ⅱ）令Cn=Sncos（anπ）（n∈N+）,求{cn}的前n项和Tn．

Answer 1

a(n)=3+(n-1)d,d>0.
s(n)=3n+n(n-1)d/2.
b(n)=q^(n-1).
12=a(2)b(2)=(3+d)q,
20=s(3)+b(2)=9+3d+q,
36=(9+3d)q.
由韦达定理知,(9+3d)和q分别为0=x^2-20x+36=(x-2)(x-18)的2个根.
因9+3d>9,因此,9+3d=18,q=2.d=3.
a(n)=3+(n-1)*3=3n,
b(n)=2^(n-1).
s(n)=3n+3n(n-1)/2.
cos[a(n)π]=cos[3nπ]
c(n)=[3n+3n(n-1)/2]cos[a(n)π]=(3/2)n(n+1)cos[3nπ]
c(2n)=(3/2)2n(2n+1)=3n(2n+1),
c(2n-1)=-(3/2)(2n-1)2n=-3n(2n-1),
c(2n-1)+c(2n)=3n,
t(2n)=3n(n+1)/2.
t(2n-1)=t(2n)-c(2n)=3n(n+1)/2 - 3n(2n+1)=3n/2.