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an}{bn}的通项（2）求求｛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an}{bn}的通项（2）求
求｛bn／2an｝的前n项和sn

Answer 1

(1)
设{an}公比为q,{bn}公差为d.an>0,则q>0
a3+b5=13
a1q²+b1+4d=13
a1=1 b1=1代入,整理,得
q²+4d=12
q²=12-4d
a5+b3=21
a1q⁴+b1+2d=21
a1=1 b1=1 q²=12-4d代入,整理,得
8d²-47d+62=0
(d-2)(8d-31)=0
d=2或d=31/8
d=2时,q²=12-4d=4 q>0 q=2
d=31/8时,q²=12-4d=-7/2(舍去)
an=a1q^(n-1)=1×2^(n-1)=2^(n-1)
bn=b1+(n-1)d=1+2(n-1)=2n-1
数列{an}的通项公式为an=2^(n-1)；数列{bn}的通项公式为bn=2n-1.
(2)
bn/(2an)=(2n-1)/2ⁿ=n/2^(n-1) -1/2ⁿ
Sn=b1/(2a1)+b2/(2a2)+...+bn/(2an)=[1/2^0+2/2+3/2²+...+n/2^(n-1)]-(1/2+1/2²+...+1/2ⁿ)
令Cn=1/2^0+2/2+3/2²+...+n/2^(n-1)
则Cn/2=1/2+2/2²+...+(n-1)/2^(n-1)+n/2ⁿ
Cn-Cn/2=Cn/2=1+1/2+...+1/2^(n-1) -n/2ⁿ
=1×(1-1/2ⁿ)/(1-1/2) -n/2ⁿ
=2-(n+2)/2ⁿ
Cn=4-(2n+4)/2ⁿ
Sn=Cn -(1/2+1/2²+...+1/2ⁿ)
=4-(2n+4)/2ⁿ -(1/2)(1-1/2ⁿ)/(1-1/2)
=3-(2n+3)/2ⁿ