设{an}是等差数列,{bn}是各项都为正数的等比数列,且a1=b1=1,a3+b35=21,a5+b3=131.求{an},{bn}的通项公式2.求数列{an/bn}的前n项和Sn

Question

设{an}是等差数列,{bn}是各项都为正数的等比数列,且a1=b1=1,a3+b35=21,a5+b3=13
1.求{an},{bn}的通项公式
2.求数列{an/bn}的前n项和Sn

Answer 1

设公差为d,公比为q,显然q>0
则2d+q^4=20 (1) 4d+q^2=12 (2)
(1)*2-(2) (2q^2+7)(q^2-4)=0
∵q>0
∴q=2 代入得d=2
1、an=1+2(n-1)=2n-1
bn=2^(n-1)
2、Sn=1+3/2+5/2^2+.+(2n-1)/2^(n-1) (3)
2Sn=2+3+5/2+.+(2n-1)/2^(n-2) (4)
(4)-(3) Sn=2+2+2/2+...+2/2^(n-2)-(2n-1)/2^(n-1)
=4+[1+1/2+...+1/2^(n-1)]-(2n-1)/2^(n-1)
=4+[1-1/2^(n-1)]/(1-1/2)-(2n-1)/2^(n-1)
=4+2-2/2^(n-1)-(2n-1)/2^(n-1)
=6-(2n+1)/2^(n-1)