Question

已知等差数列an,a3=5,a1+a2=4,数列bn的前n项和为sn,且sn=1-bn/2

Answer 1

1、设等差数列的首项为a1,公差为d,则

a3=a1+2d=5

a1+a2=2a1+d=4

∴a1=1,d=2

∴{an}的通项公式为：an=1+(n-1)*2=2n-1

∵bn的前n项和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}是首项b1=3/2 公比q=1/2的等比数列

通项公式为：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