已知数列{an}a1=1的等比数列且an>0,{bn}首相为1的等差数列,又a5+b3=21,a3+b5=131〕求数列{An}和{Bn}的通项公式;〔2〕求{Bn/2An}的前n项和Sn

Question

已知数列{an}a1=1的等比数列且an>0,{bn}首相为1的等差数列,又a5+b3=21,a3+b5=13
1〕求数列{An}和{Bn}的通项公式;
〔2〕求{Bn/2An}的前n项和Sn

Answer 1

a5+b3=21
a1q^4+b1+2d=21
q^4+1+2d=21
q^4+2d=20.1
a3+b5=13
a1q^2+b1+4d=13
q^2+1+4d=13
q^2+4d=12.2
1式*2-2式得
2q^4-q^2=28
2q^4-q^2-28=0
(2q^2+7)(q^2-4)=0
(q^2-4)=0
(q-2)(q+2)=0
q=2或q=-2(舍去）
q^4+2d=20
2^4+2d=20
2d=4
d=2
an=a1q^(n-1)
=2^(n-1)
bn=b1+(n-1)d
=1+2(n-1)
=2n-1
bn/2an
=(2n-1)/2*2^(n-1)
=(2n-1)/2^n
sn=1/2^1+3/2^2+5/2^3+.+(2n-1)/2^n
sn/2=1/2^2+3/2^3+5/2^4+.+(2n-1)/2^(n+1)
sn-sn/2=1/2^1+2/2^2+2/2^3+.+2/2^n-(2n-1)/2^(n+1)
sn/2=1/2^1+1/2^1+1/2^2+.+/2^(n-1)-(2n-1)/2^(n+1)
sn/2=1/2^1+1/2*[1-(1/2)^(n-1)]/(1-1/2)-(2n-1)/2^(n+1)
sn/2=1/2+1-(1/2)^(n-1)-(2n-1)/2^(n+1)
sn/2=3/2-1/2^(n-1)-(2n-1)/2^(n+1)
sn/2=3/2-4/2^(n+1)-(2n-1)/2^(n+1)
sn/2=3/2-(4+2n-1)/2^(n+1)
sn/2=3/2-(2n+3)/2^(n+1)
sn=3-(2n+3)/2^n