1、設(shè)等差數(shù)列的首項(xiàng)為a1,公差為d,則
a3=a1+2d=5
a1+a2=2a1+d=4
∴a1=1,d=2
∴{an}的通項(xiàng)公式為：an=1+(n-1)*2=2n-1
∵bn的前n項(xiàng)和Sn=1-bn/2
b1=S1=1-b1/2
3/2b1=1
b1=2/3
S(n-1)=1-b(n-1)/2
bn=Sn-S(n-1)=(1-bn/2)-[1-b(n-1)/2]=b(n-1)-bn
2bn=b(n-1)
bn=1/2b(n-1)
∴{bn}是首項(xiàng)b1=3/2 公比q=1/2的等比數(shù)列
通項(xiàng)公式為：bn=3/2*(1/2)^(n-1)=3*(1/2)^n
2、cn=anbn/2
=(2n-1)[3*(1/2)^n]/2
=(2n-1)[3*(1/2)^(n+1)]
=3(2n-1)*/2^(n+1)
Tn=(1*3)/2^2+(3*3)/2^3+……+3(2n-1)*/2^(n+1)
2Tn=(1*3)/2+(3*3)/2^2+……+3(2n-1)*/2^n
Tn=2Tn-Tn=3/2-6[1/2^2+1/2^3+……+1/2^(n+1)]-3(2n-1)*/2^(n+1)
=3/2-3(2n-1)*/2^(n+1)-6[1/4*[1-(1/2)^n]/(1-1/2)
=3/2-3(2n-1)*/2^(n+1)+3*[1-(1/2)^n]
=9/2-3(2n-1)*/2^(n+1)-3*(1/2)^n